🧠:Okay, so I need to find the area of the region defined by the inequality (x^{6} - x^{2} + y^{2} leq 0). Let me start by understanding what this inequality represents. It seems like a region in the xy-plane where the combination of x and y satisfies this equation. My goal is to visualize this region and then compute its area.First, let's rearrange the inequality to make it more understandable. If I move the (y^2) term to the other side, the inequality becomes:[y^{2} leq x^{2} - x^{6}]Hmm, so this tells me that (y^2) is less than or equal to (x^{2} - x^{6}). Since (y^2) is always non-negative, the right side of the inequality must also be non-negative for there to be any real solutions. Therefore, the region exists only where (x^{2} - x^{6} geq 0). Let me solve this condition first. [x^{2} - x^{6} geq 0]Factor out (x^2):[x^{2}(1 - x^{4}) geq 0]Which can be further factored as:[x^{2}(1 - x^{2})(1 + x^{2}) geq 0]Since (1 + x^{2}) is always positive (because squaring any real number gives a non-negative result, and adding 1 keeps it positive), we can ignore that term for the inequality. So the inequality reduces to:[x^{2}(1 - x^{2}) geq 0]Now, (x^{2}) is non-negative for all real x, so the sign of the entire expression depends on (1 - x^{2}). Therefore, (1 - x^{2} geq 0) because (x^{2}) is multiplied by a non-negative term. Solving (1 - x^{2} geq 0):[x^{2} leq 1 implies |x| leq 1]Therefore, the region exists only when (x) is between -1 and 1. So, the original inequality (y^{2} leq x^{2} - x^{6}) is only valid for (x in [-1, 1]). For each x in this interval, the values of y are bounded between (-sqrt{x^{2} - x^{6}}) and (sqrt{x^{2} - x^{6}}). Therefore, the region is symmetric about the x-axis because y is squared. So the area can be computed by integrating the upper half of the region and then doubling it. The area (A) would be:[A = 2 int_{-1}^{1} sqrt{x^{2} - x^{6}} , dx]But since the function inside the integral is even (symmetric about the y-axis), because replacing (x) with (-x) gives the same value, we can simplify the integral from 0 to 1 and double it:[A = 2 times 2 int_{0}^{1} sqrt{x^{2} - x^{6}} , dx = 4 int_{0}^{1} sqrt{x^{2} - x^{6}} , dx]So now, the key is to compute this integral. Let me try to simplify the integrand. The integrand is (sqrt{x^{2} - x^{6}}). Let's factor inside the square root:[sqrt{x^{2}(1 - x^{4})} = sqrt{x^{2}} sqrt{1 - x^{4}} = |x| sqrt{1 - x^{4}}]But since we're integrating from 0 to 1, (x) is non-negative, so (|x| = x). Therefore, the integrand simplifies to:[x sqrt{1 - x^{4}}]Therefore, the integral becomes:[4 int_{0}^{1} x sqrt{1 - x^{4}} , dx]Now, let's consider substitution to solve this integral. Let me set (u = x^2). Then, (du = 2x , dx), which implies that (x , dx = frac{du}{2}). Let me check:If (u = x^2), then when x = 0, u = 0, and when x = 1, u = 1. The integral becomes:[4 int_{u=0}^{u=1} sqrt{1 - u^{2}} times frac{du}{2} = 2 int_{0}^{1} sqrt{1 - u^{2}} , du]Ah, this simplifies nicely. The integral (int sqrt{1 - u^{2}} , du) is a standard integral, which is related to the area of a semicircle. Specifically, the integral from 0 to 1 of (sqrt{1 - u^2} , du) is equal to (pi/4), because it's the area of a quarter of a unit circle. Wait, let me confirm that. The integral from 0 to 1 of (sqrt{1 - u^2} , du) is indeed the area of the first quadrant portion of the unit circle, which is (pi/4). Therefore:[2 times frac{pi}{4} = frac{pi}{2}]Therefore, the area (A) is (pi/2).But let me verify each step carefully to make sure I haven't made a mistake. First, original inequality: (x^6 - x^2 + y^2 leq 0), which rearranged gives (y^2 leq x^2 - x^6). Correct. Then, considering the right-hand side (x^2 - x^6 geq 0), factoring gives (x^2(1 - x^4) geq 0). Since (x^2 geq 0), then (1 - x^4 geq 0) which implies (x^4 leq 1) so (|x| leq 1). Correct. Thus, the bounds on x are from -1 to 1, and for each x in that interval, y ranges between (pm sqrt{x^2 - x^6}). Then, since the region is symmetric about both the x-axis and y-axis, we can compute the area in the first quadrant and multiply by 4. Wait, hold on. Wait, actually, the original integral was set up as doubling the integral from -1 to 1 because of the symmetry about the x-axis. But actually, since the function is even in x (symmetric about the y-axis), and also the y term is squared, so symmetric about the x-axis. Therefore, the entire region is symmetric in all four quadrants? Wait, no. Let's see.Wait, if we have (y^2 leq x^2 - x^6), then for each x between -1 and 1, y can be positive or negative, so symmetric about the x-axis. However, the expression (x^2 - x^6) is even in x, since replacing x with -x gives the same value. Therefore, the region is also symmetric about the y-axis. Therefore, the entire region is symmetric in all four quadrants. Therefore, we can compute the area in the first quadrant (where x ≥ 0 and y ≥ 0) and multiply by 4. But in my initial setup, I considered integrating from -1 to 1 and doubling because of the x-axis symmetry, but actually, since it's symmetric about both axes, integrating from 0 to 1 and multiplying by 4 would be more efficient. However, let's check the original steps. The user wrote:Original integral after simplifying symmetry:(A = 2 int_{-1}^{1} sqrt{x^{2} - x^{6}} , dx)But then recognized the integrand is even, so (2 times 2 int_{0}^{1} sqrt{x^{2} - x^{6}} , dx = 4 int_{0}^{1} sqrt{x^{2} - x^{6}} , dx). That seems correct. Because initially, the area is twice (due to upper and lower halves) the integral from -1 to 1, but since the integrand is even, that's 2 times 2 times integral from 0 to 1. So total 4 times. Correct.Then, simplifying the integrand to (x sqrt{1 - x^4}). Then substitution (u = x^2), which leads to (du = 2x dx), so (x dx = du/2), leading to:(4 times frac{1}{2} int_{0}^{1} sqrt{1 - u^2} du = 2 int_{0}^{1} sqrt{1 - u^2} du). Then, since the integral of sqrt(1 - u^2) from 0 to 1 is pi/4, so 2*(pi/4) = pi/2. Wait, but let me confirm the substitution again. Starting with:Original integral after substitution:4 * integral from 0 to1 of x*sqrt(1 - x^4) dxSet u = x^2, so du = 2x dx → x dx = du/2Then sqrt(1 - x^4) = sqrt(1 - (x^2)^2) = sqrt(1 - u^2)Therefore, the integral becomes 4 * integral from u=0 to u=1 of sqrt(1 - u^2) * (du/2) = 2 * integral from 0 to1 sqrt(1 - u^2) du. Yes, correct.Then, the integral of sqrt(1 - u^2) du from 0 to1 is indeed pi/4. Because the integral from 0 to1 sqrt(1 - u^2) du is the area of a quarter of the unit circle in the first quadrant, which is (pi*(1)^2)/4 = pi/4. Therefore, multiplying by 2 gives pi/2. Therefore, the area is pi/2. But let me make sure that there isn't a miscalculation here. Let's recall that the integral of sqrt(a^2 - u^2) du from 0 to a is (pi a^2)/4, which in this case a=1, so pi/4. Correct.Therefore, the area should be pi/2.Wait, but let me check with another substitution to verify. Suppose instead of substituting u = x^2, we try a trigonometric substitution. Let's see.Let me consider the integral (int sqrt{1 - u^2} du). If we set u = sinθ, then du = cosθ dθ, and sqrt(1 - u^2) = cosθ. Then the integral becomes:∫ cosθ * cosθ dθ = ∫ cos^2θ dθ. Using the identity cos^2θ = (1 + cos2θ)/2, the integral becomes:(θ/2) + (sin2θ)/4 + CChanging back to u, since u = sinθ, θ = arcsin(u), and sin2θ = 2 sinθ cosθ = 2u sqrt(1 - u^2). Therefore, the integral is:(arcsin(u)/2) + (u sqrt(1 - u^2))/2 + CEvaluating from 0 to1:At u=1: arcsin(1)/2 + (1 * 0)/2 = (pi/2)/2 + 0 = pi/4At u=0: arcsin(0)/2 + 0 = 0Therefore, the integral from 0 to1 is pi/4. Therefore, 2*(pi/4) = pi/2. Correct.Therefore, the area is indeed pi/2. But let me check with another approach. Suppose we parameterize the curve.The original inequality is (x^6 - x^2 + y^2 leq 0). Let's see if this represents a closed curve. For x in [-1,1], the equation (y = pm sqrt{x^2 - x^6}) defines the boundary. Plotting this might help visualize. At x=0, y=0. At x=1, x^6 - x^2 = 1 -1 =0, so y=0. Similarly at x=-1. In between, the term x^2 -x^6 is positive. Let's check at x=1/2:(1/2)^2 - (1/2)^6 = 1/4 - 1/64 = 16/64 - 1/64 = 15/64, which is positive. So the curve is a closed loop between x=-1 and x=1, symmetric across both axes. Since we computed the area via integration and arrived at pi/2, but let me check if there is a geometric interpretation. Wait, the integral after substitution became 2 * integral from 0 to1 sqrt(1 - u^2) du. That is, 2 times a quarter of the area of the unit circle, which is 2*(pi/4) = pi/2. Therefore, the area inside the given inequality is pi/2. Alternatively, if I consider polar coordinates. Let me try converting the equation to polar coordinates to see if that's another way to compute the area. In polar coordinates, x = r cosθ, y = r sinθ. Then, substitute into the inequality:x^6 - x^2 + y^2 ≤ 0So:(r cosθ)^6 - (r cosθ)^2 + (r sinθ)^2 ≤ 0Simplify:r^6 cos^6θ - r^2 cos^2θ + r^2 sin^2θ ≤ 0Factor out r^2:r^2 [ r^4 cos^6θ - cos^2θ + sin^2θ ] ≤ 0Since r^2 is non-negative, the inequality reduces to:r^4 cos^6θ - cos^2θ + sin^2θ ≤ 0But this seems complicated. Let me see if we can express this in terms of sin^2θ and cos^2θ.Note that sin^2θ = 1 - cos^2θ. Therefore:r^4 cos^6θ - cos^2θ + (1 - cos^2θ) ≤ 0Simplify:r^4 cos^6θ - 2 cos^2θ + 1 ≤ 0This equation may not be straightforward to solve for r. It might not lead to an easier integral in polar coordinates. Therefore, perhaps sticking with the Cartesian approach is better. Alternatively, perhaps parametrizing using u = x^2 as before. Wait, we already did substitution and got the answer. So unless there's an error in that substitution, the answer should be correct. Another check: let's compute the integral numerically. Let's approximate the integral (4 int_{0}^{1} x sqrt{1 - x^4} dx). Take x=0, the integrand is 0. At x=1, the integrand is 0. The function x*sqrt(1 - x^4) reaches a maximum somewhere in between. Let's approximate the integral using, say, the midpoint rule with a few intervals.But this might be time-consuming, but just for a rough check. Alternatively, use substitution t = x^2, so x = sqrt(t), dx = (1)/(2 sqrt(t)) dt. Then the integral becomes:4 * ∫_{0}^{1} sqrt(t) * sqrt(1 - t^2) * (1/(2 sqrt(t))) dt = 4 * (1/2) ∫_{0}^{1} sqrt(1 - t^2) dt = 2 * (pi/4) = pi/2, same result. So this confirms the substitution again.Therefore, I think the answer is indeed pi/2.But let me consider another perspective. Suppose we let t = x^3. Wait, but not sure. Alternatively, is there a connection to the area of a circle or ellipse?Wait, the equation after substitution became similar to a circle equation. In the u substitution, the integral became 2 * integral of sqrt(1 - u^2) du, which is like a semicircle. But why?Alternatively, considering the original inequality: y^2 ≤ x^2 - x^6. If we rewrite this as:y^2 + x^6 ≤ x^2But not sure. It's a higher-degree equation, so not a standard conic section. Alternatively, parametrize x in terms of sinθ or cosθ. Let me try x^2 = sinθ, but perhaps not useful. Alternatively, set x^2 = t, so t ∈ [0,1], then y^2 ≤ t - t^3. Then the area would be 2 * ∫_{-1}^{1} sqrt(t - t^3) dx, but x is in terms of t. Wait, maybe complicating.Alternatively, think of the curve as y = ±sqrt(x^2 -x^6). For each x, the y extends from -sqrt(x^2 -x^6) to sqrt(x^2 -x^6). Therefore, the area is 2 * ∫_{-1}^{1} sqrt(x^2 -x^6) dx, as before. Then, as established, using substitution, it becomes pi/2.Alternatively, check with specific x values. For example, when x=0, the y range is 0. When x approaches 1, y approaches 0. The maximum y occurs where x^2 -x^6 is maximized. Take derivative of x^2 -x^6 with respect to x: 2x -6x^5. Set to zero: 2x -6x^5 =0 → x(2 -6x^4)=0. Solutions at x=0 and 2 -6x^4=0 → x^4=1/3 → x=(1/3)^{1/4} ≈ 0.7598. At this x, the maximum y is sqrt((1/3) - (1/3)^3) = sqrt((1/3) - (1/27)) = sqrt(8/27) ≈ 0.544. So the curve has a maximum around x≈0.76, y≈0.54. But regardless of the shape, the area computation via substitution seems valid, leading to pi/2. Another way to think: the integral we computed is 4 times the integral from 0 to1 of x*sqrt(1 -x^4) dx. Let me consider substitution z = x^2. Then dz = 2x dx → x dx = dz/2. But this is similar to previous substitution. So when x=0, z=0; x=1, z=1. Integral becomes 4 * ∫0^1 sqrt(1 - z^2) * (dz/2) = 2 ∫0^1 sqrt(1 - z^2) dz = same as before, pi/2. Thus, all substitution methods lead to the same result. Therefore, after thorough analysis and multiple checks, I can confirm that the area of the region defined by (x^{6} - x^{2} + y^{2} leq 0) is (pi/2).Final AnswerThe area of the region is boxed{dfrac{pi}{2}}.

✅:To find the area of the region defined by the inequality (x^6 - x^2 + y^2 leq 0), we start by rearranging the inequality:[y^2 leq x^2 - x^6]This implies that (x^2 - x^6 geq 0). Factoring this expression, we get:[x^2(1 - x^4) geq 0]Since (x^2 geq 0) for all real (x), the inequality reduces to (1 - x^4 geq 0), which gives (|x| leq 1). Therefore, the region exists for (x in [-1, 1]).The area can be computed by integrating the upper half of the region and then doubling it. Due to symmetry about both the x-axis and y-axis, we can simplify the integral by considering the first quadrant and multiplying by 4:[A = 4 int_{0}^{1} sqrt{x^2 - x^6} , dx]Simplifying the integrand:[sqrt{x^2 - x^6} = x sqrt{1 - x^4}]Using the substitution (u = x^2), we get (du = 2x , dx) or (x , dx = frac{du}{2}). The integral becomes:[4 int_{0}^{1} x sqrt{1 - x^4} , dx = 2 int_{0}^{1} sqrt{1 - u^2} , du]The integral (int_{0}^{1} sqrt{1 - u^2} , du) represents the area of a quarter of the unit circle, which is (frac{pi}{4}). Therefore:[2 times frac{pi}{4} = frac{pi}{2}]Thus, the area of the region is:[boxed{dfrac{pi}{2}}]

🔑:To find the area of the region in the ( xy )-plane defined by the inequality ( x^6 - x^2 + y^2 leq 0 ), we follow these steps:1. Rewrite the given inequality: [ x^6 - x^2 + y^2 leq 0 quad Rightarrow quad y^2 leq x^2 - x^6 ] This implies: [ |y| leq sqrt{x^2 - x^6} ]2. Determine the boundaries of the region: The right-hand side, ( sqrt{x^2 - x^6} ), is 0 when: [ x^2 - x^6 = 0 quad Rightarrow quad x^2(1 - x^4) = 0 ] Thus, ( x = -1, 0, 1 ). This region is symmetric about the ( x )- and ( y )-axes.3. Set up the integral to calculate the area: Given the symmetry, we can compute the area in the first quadrant and multiply by 4: [ A = 4 int_0^1 sqrt{x^2 - x^6} , dx ]4. Simplify the integrand: Notice that: [ sqrt{x^2 - x^6} = x sqrt{1 - x^4} ] Thus: [ A = 4 int_0^1 x sqrt{1 - x^4} , dx ]5. Use substitution to evaluate the integral: Let ( u = x^2 ). Then ( du = 2x , dx ), and the limits of integration change from ( x ) in ([0, 1]) to ( u ) in ([0, 1]): [ A = 4 int_0^1 x sqrt{1 - x^4} , dx = 4 int_0^1 sqrt{1 - u^2} cdot frac{1}{2} , du = 2 int_0^1 sqrt{1 - u^2} , du ]6. Recognize the integral as a known form: The integral ( int_0^1 sqrt{1 - u^2} , du ) is the area of a quarter circle of radius 1: [ 2 int_0^1 sqrt{1 - u^2} , du = 2 cdot frac{pi}{4} = frac{pi}{2} ]7. Conclusion: Hence, the area of the region is: [ boxed{frac{pi}{2}} ]

🧠:Okay, let's try to figure out this problem. So, there's a square plot, right? And it was expanded into a new square. The new square has a side that's 5 meters longer than the original. The area increased by 225 square meters. We need to find the area of the original plot.Hmm, let's start by visualizing this. Imagine a square, let's call the original side length 's'. So, the original area is s squared, which is s². Then, they expanded it to a new square where each side is 5 meters longer. So, the new side length is s + 5. The area of the new square would then be (s + 5)².The problem says the area increased by 225 square meters. That means the new area minus the original area is 225. So, mathematically, that would be (s + 5)² - s² = 225.Let me write that equation down:(s + 5)² - s² = 225Now, let's expand the left side. (s + 5)² is s² + 10s + 25. So, subtracting s² from that gives:s² + 10s + 25 - s² = 10s + 25So, the equation simplifies to:10s + 25 = 225Now, solving for s:10s = 225 - 2510s = 200s = 200 / 10s = 20 metersSo, the original side length is 20 meters. Therefore, the original area is s², which is 20² = 400 square meters.Wait, let me check that again to make sure I didn't make a mistake. Let's plug the numbers back into the original equation.Original area: 20² = 400New side length: 20 + 5 = 25 metersNew area: 25² = 625Difference in area: 625 - 400 = 225Yes, that matches the problem statement. So, the calculations seem correct.Alternatively, I can think of the area increase as the area of the added parts. When you expand a square by 5 meters on each side, the increase in area isn't just a strip on one side but on all four sides. However, visualizing that might be a bit more complex. Let me try that approach too to confirm.Expanding the square by 5 meters on each side adds a border around the original square. This border consists of four rectangles: one on each side of the original square. However, if you just think of adding 5 meters to each side, the corners where the new additions meet will overlap. Each corner is a 5x5 square. So, the total added area can be broken down into:- Two rectangles on the top and bottom: each has a length of the original side 's' and a width of 5. So, each of these has an area of 5s. Since there are two, that's 10s.- Two rectangles on the sides: each has a length of 5 (the width) and a height of the original side plus 5 (because when you add to the side, the height includes the original side plus the new addition). Wait, no. Wait, actually, when you add 5 meters to each side, the side additions are next to the original square, so each side rectangle would be 5 meters in width and the same height as the original square. Wait, maybe I'm complicating it.Alternatively, when you add 5 meters to each side, the total expansion can be thought of as the original square plus an additional border. The border's area is the difference between the new area and the original area, which is 225. If I break down the border:- Each side of the original square gets extended by 5 meters. So, the four sides would each have a rectangle of area s * 5. That's 4 * 5s = 20s. However, this counts the four corners twice because each corner is part of two sides. The four corners are each 5x5 squares, so their total area is 4 * 25 = 100. Therefore, the actual added area is 20s - 100 + 100? Wait, no. Wait, when you add 5 meters to each side, the total area added is 5*(s + 5) + 5*(s) but I think I need to visualize it better.Wait, perhaps another way. If you have a square with side s, and you add 5 meters to each side, making the new side s + 5. The added area is the area of the new square minus the original. So, (s + 5)^2 - s^2 = 225. Which simplifies to 10s + 25 = 225, leading to s = 20. So that's the same result. Therefore, the original area is 400.Alternatively, if I think of the expansion as adding a 5-meter strip to one side, but since it's a square, you have to add it to all sides. But no, actually, when you make a square larger by 5 meters on each side, you're adding 5 meters to both ends of each side. So, each side's length increases by 5 meters on both ends, which totals 10 meters. Wait, no. Wait, if the original side is s, and you add 5 meters to each side, meaning each side is extended by 5 meters in both directions. So, for example, the left side is extended 5 meters to the left, and the right side 5 meters to the right, so the total new side length is s + 5 + 5 = s + 10. Wait, hold on. Wait, maybe I'm getting confused here.Wait, no, the problem states that the new square has a side that is 5 meters longer than the original. So, it's not adding 5 meters to each side of the original square, but rather each side of the new square is 5 meters longer than the original square's side. So, if the original side is s, the new side is s + 5. Therefore, the increase in area is (s + 5)^2 - s^2 = 225. So, that's the correct equation. So, the previous calculation is correct. The confusion earlier was perhaps due to misinterpreting how the side length is increased. But according to the problem statement, the new square's side is 5 meters longer, not adding 5 meters to each side. So, the side length increases by 5, leading to s + 5. Therefore, the area difference is as calculated. So, the original area is 400, which checks out when plugging back into the equation.Another way to verify is to consider specific numbers. Suppose the original side was 20, adding 5 gives 25, area difference is 625 - 400 = 225, which matches. If the original side was, say, 10, then new side would be 15, area difference 225 - 100 = 125, which is less than 225. So, 20 seems correct.Alternatively, maybe there's a different approach using algebra tiles or geometric visualization, but the algebraic method seems straightforward and correct here. I don't see any mistakes in the steps. So, the answer should be 400 square meters.

✅:The original square plot had a side length of ( s ) meters. When expanded, the new side length becomes ( s + 5 ) meters. The increase in area is given as 225 m². Step-by-Step Solution:1. Original Area: ( s^2 )2. New Area: ( (s + 5)^2 )3. Area Difference Equation: [ (s + 5)^2 - s^2 = 225 ]4. Expand and Simplify: [ s^2 + 10s + 25 - s^2 = 225 implies 10s + 25 = 225 ]5. Solve for ( s ): [ 10s = 200 implies s = 20 text{ meters} ]6. Original Area: [ s^2 = 20^2 = 400 text{ m²} ]Verification: New side = 25 m, New area = ( 25^2 = 625 ) m². Difference = ( 625 - 400 = 225 ) m², confirming the solution is correct.Answer: The area of the original plot is (boxed{400}) square meters.

🔑:1. Let ( s ) be the side length of the original square.2. The side length of the new square is ( s + 5 ) meters.3. The area of the original square is ( s^2 ).4. The area of the new square is ( (s + 5)^2 ).5. The problem states that the area has increased by 225 m(^2). Thus, we set up the equation: [ (s + 5)^2 = s^2 + 225 ]6. Expand the left-hand side: [ s^2 + 10s + 25 = s^2 + 225 ]7. Subtract ( s^2 ) from both sides to simplify: [ 10s + 25 = 225 ]8. Subtract 25 from both sides: [ 10s = 200 ]9. Divide by 10 to solve for ( s ): [ s = 20 ]10. Calculate the area of the original square: [ s^2 = 20^2 = 400 ]Conclusion:[boxed{400}]

🧠:Okay, let's try to tackle this problem. So, we have a sequence {a_n} that follows the recurrence relation a_n = a_{n-1} - a_{n-2} for n ≥ 3. We’re told that the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492. We need to find the sum of the first 2001 terms. Hmm, interesting. Let me start by understanding the sequence better.First, since the recurrence relation is a linear recurrence, maybe the sequence is periodic? That often happens with such recurrences. Let me check that. Let's try to compute the first few terms in terms of a1 and a2. Let's say a1 = x and a2 = y. Then:a3 = a2 - a1 = y - xa4 = a3 - a2 = (y - x) - y = -xa5 = a4 - a3 = (-x) - (y - x) = -ya6 = a5 - a4 = (-y) - (-x) = x - ya7 = a6 - a5 = (x - y) - (-y) = xa8 = a7 - a6 = x - (x - y) = yWait, a7 and a8 are back to x and y, which are a1 and a2. So this sequence has a period of 6. That is, every 6 terms, the sequence repeats. Let me confirm that:a1 = xa2 = ya3 = y - xa4 = -xa5 = -ya6 = x - ya7 = x (same as a1)a8 = y (same as a2)Yep, so the period is 6. Therefore, the sequence is periodic with period 6. That's a key insight. So the sum over any 6 consecutive terms should be the same. Let's calculate the sum of one period:Sum = a1 + a2 + a3 + a4 + a5 + a6Substituting the terms:Sum = x + y + (y - x) + (-x) + (-y) + (x - y)Let me compute this step by step:x + y = x + yAdding (y - x): x + y + y - x = 2yAdding (-x): 2y - xAdding (-y): 2y - x - y = y - xAdding (x - y): y - x + x - y = 0So the sum over each period of 6 terms is 0. That's important. So every 6 terms, the sum cancels out. Therefore, when summing a large number of terms, the total sum will depend only on the remainder when the number of terms is divided by 6.For example, if we have N terms, then N = 6*q + r, where r is 0,1,2,3,4,5. Then the sum S(N) = sum_{n=1}^N a_n = sum_{n=1}^r a_n, because the sum of each full period is zero.Therefore, the sum S(N) is equal to the sum of the first r terms, where r = N mod 6.Given that, let's denote S(N) = sum_{n=1}^r a_n, where r = N mod 6. So for N ≥ 6, S(N) is just the sum of the first r terms.So in the problem, we are given S(1492) = 1985 and S(1985) = 1492. We need to find S(2001).First, let's compute 1492 mod 6, 1985 mod 6, and 2001 mod 6 to find r for each.Calculating 1492 divided by 6:6*248 = 1488, so 1492 - 1488 = 4. So 1492 mod 6 = 4. Therefore, S(1492) = sum_{n=1}^4 a_n = 1985.Similarly, 1985 divided by 6:6*330 = 1980, so 1985 - 1980 = 5. So 1985 mod 6 = 5. Therefore, S(1985) = sum_{n=1}^5 a_n = 1492.And 2001 divided by 6:6*333 = 1998, so 2001 - 1998 = 3. So 2001 mod 6 = 3. Therefore, S(2001) = sum_{n=1}^3 a_n.So, our goal is to find sum_{n=1}^3 a_n, which is a1 + a2 + a3. Let's denote this as S3.But we need to express S3 in terms of the given sums S(1492) and S(1985). Let's note that S(1492) = sum_{1}^4 a_n = 1985, and S(1985) = sum_{1}^5 a_n = 1492.Let me write down the equations:Equation 1: a1 + a2 + a3 + a4 = 1985Equation 2: a1 + a2 + a3 + a4 + a5 = 1492If we subtract Equation 1 from Equation 2, we get:a5 = 1492 - 1985 = -493So, a5 = -493But we can express a5 in terms of the earlier terms. Let's recall the sequence:a1 = xa2 = ya3 = y - xa4 = -xa5 = -ya6 = x - ySo, a5 = -y. Therefore, from a5 = -493, we have:-y = -493 ⇒ y = 493So, a2 = y = 493Then, let's find a4. From the sequence, a4 = -x. From Equation 1:a1 + a2 + a3 + a4 = x + y + (y - x) + (-x) = x + y + y - x - x = (x - x - x) + (y + y) = (-x) + 2ySo, -x + 2y = 1985We already know y = 493. Substitute:-x + 2*493 = 1985 ⇒ -x + 986 = 1985 ⇒ -x = 1985 - 986 ⇒ -x = 999 ⇒ x = -999Therefore, a1 = x = -999So, now we can compute all the terms:a1 = -999a2 = 493a3 = a2 - a1 = 493 - (-999) = 493 + 999 = 1492a4 = -a1 = -(-999) = 999a5 = -a2 = -493a6 = a3 - a4 = 1492 - 999 = 493Wait, a6 is 493? Let me check:Wait, a3 is 1492, a4 is 999, so a5 is a4 - a3 = 999 - 1492 = -493. Wait, but the recurrence is a_n = a_{n-1} - a_{n-2}, so for n=5: a5 = a4 - a3 = 999 - 1492 = -493. Correct.Then a6 = a5 - a4 = (-493) - 999 = -1492. Wait, hold on, this contradicts what I had earlier. Wait, perhaps my initial calculation of a6 was wrong. Let's recast the terms step by step:a1 = -999a2 = 493a3 = a2 - a1 = 493 - (-999) = 1492a4 = a3 - a2 = 1492 - 493 = 999Wait, hold on. Wait, the recurrence is a_n = a_{n-1} - a_{n-2}. So for a3, it's a2 - a1. For a4, it's a3 - a2. For a5, it's a4 - a3. For a6, it's a5 - a4. Etc.Wait, earlier, when I derived the terms, I thought a4 was -x. But maybe that was under the assumption of a different starting point? Let me check again.Wait, in the initial step-by-step, we had:a1 = xa2 = ya3 = y - xa4 = a3 - a2 = (y - x) - y = -xa5 = a4 - a3 = (-x) - (y - x) = -ya6 = a5 - a4 = (-y) - (-x) = x - yThen a7 = a6 - a5 = (x - y) - (-y) = xWhich is correct, so if we follow that structure, then:a1 = x = -999a2 = y = 493a3 = y - x = 493 - (-999) = 1492a4 = -x = -(-999) = 999a5 = -y = -493a6 = x - y = (-999) - 493 = -1492Then a7 = a6 - a5 = (-1492) - (-493) = -1492 + 493 = -999, which is a1a8 = a7 - a6 = (-999) - (-1492) = -999 + 1492 = 493 = a2So the sequence is periodic with period 6 as expected. Therefore, the terms are:1: -9992: 4933: 14924: 9995: -4936: -14927: -9998: 493...and so on.Wait, but earlier, when I tried to compute a6 as a5 - a4, that would be (-493) - 999 = -1492, which is correct. So in the first period:Sum of 6 terms: (-999) + 493 + 1492 + 999 + (-493) + (-1492) = Let's compute:Start adding up:-999 + 493 = -506-506 + 1492 = 986986 + 999 = 19851985 - 493 = 14921492 - 1492 = 0Yes, sum is 0. So that's correct.Now, going back to the problem. We found a1 = -999, a2 = 493, a3 = 1492, a4 = 999, a5 = -493, a6 = -1492, and then it repeats.Given that, we can compute the sum of the first 4 terms (since 1492 mod 6 = 4), which is S(1492) = a1 + a2 + a3 + a4 = -999 + 493 + 1492 + 999. Let's compute that:-999 + 493 = -506-506 + 1492 = 986986 + 999 = 1985. Correct, that's given.Similarly, sum of first 5 terms (1985 mod 6 =5) is S(1985) = a1 + a2 + a3 + a4 + a5 = 1985 + (-493) = 1492. Correct.So now, we need to find the sum of the first 2001 terms. As 2001 mod 6 = 3 (since 2001 divided by 6 is 333*6=1998, 2001-1998=3). So sum is a1 + a2 + a3.Compute a1 + a2 + a3:-999 + 493 + 1492.Compute step by step:-999 + 493 = -506-506 + 1492 = 986Wait, but that's 986. But let me check:Wait, a1 is -999, a2 is 493, a3 is 1492.Sum: -999 + 493 = -506; -506 + 1492 = 986. So the sum of the first 3 terms is 986. But is this correct? Let's verify.Wait, but earlier, when we found the sum of the first 4 terms (1985), that was sum of first 3 terms plus a4 (which is 999). So 986 + 999 = 1985, which matches. So that's correct. Therefore, the sum of the first 3 terms is 986, so the sum of the first 2001 terms is 986.But let me just make sure there's no mistake here. Let me go through the steps again.We found that the sum of the first N terms is equal to the sum of the first r terms where r = N mod 6.Given that, 2001 mod 6 is 3. So sum is a1 + a2 + a3 = -999 + 493 + 1492 = 986.But let me check if we did everything correctly. Let's confirm the initial conditions again.Given S(1492) = 1985 and S(1985) = 1492.We found:a1 = -999, a2 = 493, a3 = 1492, a4 = 999, a5 = -493, a6 = -1492Sum of first 4 terms: -999 + 493 + 1492 + 999 = 1985 ✔️Sum of first 5 terms: 1985 + (-493) = 1492 ✔️Therefore, the values are consistent. Therefore, the sum of first 3 terms is indeed 986, which is S(2001).Wait, but before finalizing, let me just check 2001 mod 6 again.Divide 2001 by 6:6*333=19982001-1998=3. So 2001 mod 6 is 3. So sum is first 3 terms. Correct.Therefore, the answer should be 986. Therefore, the sum of the first 2001 terms is 986. So the final answer is boxed{986}.Final Answerboxed{986}

✅:Given a sequence ({a_n}) that satisfies the recurrence relation (a_n = a_{n-1} - a_{n-2}) for (n geq 3), we need to find the sum of the first 2001 terms. The sequence is known to have a period of 6, meaning it repeats every 6 terms. Key steps:1. Periodicity Analysis: The sequence ({a_n}) is periodic with period 6. The terms are: [ begin{aligned} a_1 &= x, a_2 &= y, a_3 &= y - x, a_4 &= -x, a_5 &= -y, a_6 &= x - y. end{aligned} ] The sum of one period (6 terms) is 0.2. Modulo Calculations: - (1492 mod 6 = 4) - (1985 mod 6 = 5) - (2001 mod 6 = 3)3. Sum Calculations: - The sum of the first 1492 terms (4 terms modulo 6) is given as 1985. - The sum of the first 1985 terms (5 terms modulo 6) is given as 1492.4. Solving for Initial Terms: - From the sum of the first 4 terms: (a_1 + a_2 + a_3 + a_4 = 1985) - From the sum of the first 5 terms: (a_1 + a_2 + a_3 + a_4 + a_5 = 1492) - Subtracting these equations gives (a_5 = -493), leading to (a_2 = 493). - Solving for (a_1) using the sum of the first 4 terms gives (a_1 = -999).5. Term Values: - (a_1 = -999) - (a_2 = 493) - (a_3 = 1492) - (a_4 = 999) - (a_5 = -493) - (a_6 = -1492)6. Final Sum Calculation: - The sum of the first 3 terms (2001 modulo 6) is: [ a_1 + a_2 + a_3 = -999 + 493 + 1492 = 986 ]Thus, the sum of the first 2001 terms is (boxed{986}).

🔑:Given the sequence left{a_nright} which satisfies (a_n = a_{n-1} - a_{n-2}) for (n geq 3), we need to find the sum of its first 2001 terms given the sums up to 1492 and 1985. Let's detail the reference solution step by step:1. Analyze the sequence pattern: Starting with initial terms (a_n), (a_{n+1}), let's generate the next several terms: [ begin{aligned} a_{n+2} &= a_{n+1} - a_n, a_{n+3} &= a_{n+2} - a_{n+1} &= (a_{n+1} - a_n) - a_{n+1} &= -a_n, a_{n+4} &= a_{n+3} - a_{n+2} &= -a_n - (a_{n+1} - a_n) &= -a_{n+1}, a_{n+5} &= a_{n+4} - a_{n+3} &= -a_{n+1} - (-a_n) &= a_n - a_{n+1}, a_{n+6} &= a_{n+5} - a_{n+4} &= (a_n - a_{n+1}) - (-a_{n+1}) &= a_n. end{aligned} ] The sequence repeats every 6 terms: (a_n, a_{n+1}, a_{n+2}, -a_n, -a_{n+1}, a_n - a_{n+1}).2. Sum of one period (six terms): [ a_n + a_{n+1} + a_{n+2} + (-a_n) + (-a_{n+1}) + (a_n - a_{n+1}) = 0 ] The sum of any six consecutive terms in the sequence is zero.3. Define ( S_n ) for the sum of the first ( n ) terms: Since the sequence has a repeating pattern every six terms, knowing the positions of 1492 and 1985 modulo 6 is crucial: [ 1492 = 6 times 248 + 4,quad 1985 = 6 times 330 + 5 ] Thus, [ S_{1492} = a_1 + a_2 + a_3 + a_4, ] [ S_{1985} = a_1 + a_2 + a_3 + a_4 + a_5. ] We know: [ S_{1492} = 1985 quad text{and} quad S_{1985} = 1492 ] Hence, we have: [ begin{aligned} 1985 &= a_1 + a_2 + (a_1 - a_2) + (-a_1) &= 2a_2 - a_1, end{aligned} ] [ begin{aligned} 1492 &= a_1 + a_2 + (a_1 - a_2) + (-a_1) - a_{n+1} + a_n - a_{n+1} &= a_2 - a_1. end{aligned} ]4. Solve for ( a_1 ) and ( a_2 ): [ begin{aligned} 1985 &= 2a_2 - a_1, 1492 &= a_2 - a_1, end{aligned} ] To solve these simultaneously, add the second equation to twice the second one: [ 2(1492 - a_1) + a_1 = 1985, ] [ 2984 - a_1 = 1985, ] [ -a_1 = -999 quad implies quad a_1 = -999. ] Substitute (a_1) in second equation: [ 1492 = a_2 - (-999), 1492 = a_2 + 999, a_2 = 1492 - 999, a_2 = 493. ]5. Find the sum of the first 2001 terms: [ 2001 = 6 times 333 + 3, ] So, [ S_{2001} = a_1 + a_2 + (a_2 - a_1). ] Substituting values gives: [ a_1 = -999, a_2 = 493, ] [ begin{aligned} S_{2001} &= a_1 + 2a_2 - a_1 &= 2a_2 &= 2 times 493, &= 986. end{aligned} ]6. Conclusion: [ boxed{986} ]

🧠:Okay, so I need to figure out how many of these three friends are always right, given their statements and the conditions. Let's start by breaking down the problem step by step.First, each friend is either a truth-teller (always right) or a liar (always wrong). The key here is that at least one is truthful and at least one is lying. So we can't have all three being truth-tellers or all three being liars. That narrows down the possibilities.Let me note down each friend's statement:1. Friend A says: "None of us have seen this movie."2. Friend B says: "I have seen this movie, but both of you have not."3. Friend C says: "I have seen this movie."We need to determine the number of truth-tellers among them. Let's denote Friend A, B, C as A, B, C respectively. Each can be either T (truth-teller) or F (liar). We need to check all possible combinations that satisfy the given condition (at least one T and at least one F).There are 2^3 = 8 possible combinations. But considering the condition, we can eliminate the cases where all are T or all are F. That leaves 6 possible cases. Let me list them and check each one.Case 1: A=T, B=T, C=F- If A is T, then "None of us have seen the movie" is true. So A, B, C all have not seen the movie.- But B says "I have seen it, but both of you have not." Since B is T here, this would mean B has seen it, but A and C haven't. But if A is T, then B hasn't seen it. Contradiction. So this case is impossible.Case 2: A=T, B=F, C=T- A=T: None have seen the movie. So B and C haven't seen it.- B=F: His statement is false. His statement is "I have seen it, and both of you haven't." For this to be false, either he hasn't seen it, or at least one of A or C has seen it. But since A is T, none have seen it, so B hasn't seen it. Therefore, his statement is false because he claims to have seen it when he hasn't. That works.- C=T: C says "I have seen it." But according to A, C hasn't. So if C is T, then C must have seen it, which contradicts A's statement. Therefore, this case is impossible because A and C can't both be T here.Case 3: A=T, B=F, C=F- A=T: None have seen the movie.- B=F: His statement is false. So either he hasn't seen it, or at least one of A or C has. Since A is T, none have seen it, so B's statement is false because he claims to have seen it. So that's consistent.- C=F: C's statement "I have seen it" is false, so C hasn't. That's consistent with A's statement.- So this case is possible. Here, A is T, B and C are F. So one truth-teller.Case 4: A=F, B=T, C=T- A=F: His statement "None have seen it" is false. So at least one person has seen it.- B=T: "I have seen it, and both of you haven't." So B has seen it, A and C haven't. But C is T here, so C's statement "I have seen it" would be true, meaning C has seen it. But B says C hasn't. Contradiction. Therefore, this case is impossible.Case 5: A=F, B=T, C=F- A=F: So at least one has seen it.- B=T: B has seen it, and A and C haven't.- C=F: C's statement "I have seen it" is false, so C hasn't. This is consistent with B's statement.- Now check A's statement. Since at least one (B) has seen it, A's statement is false, which is correct because A is F here. So this case works. Here, B is T, A and C are F. So one truth-teller.Case 6: A=F, B=F, C=T- A=F: So at least one has seen it.- B=F: His statement is "I have seen it, and both of you haven't." For this to be false, either he hasn't seen it, or at least one of A or C has. Since A is F, but A's statement is "none have seen it," which is false, so someone has. But B's statement being false could mean he hasn't seen it, or someone else has. C is T here, so C's statement is true: C has seen it. Therefore, B's statement that both A and C haven't is false because C has. So B's statement is false, which fits.- So in this case, C is T, A and B are F. Then, A=F (so at least one has seen it, which is C), B=F (he hasn't seen it, and C has, so his statement is false), C=T (C has seen it). This works. So here, one truth-teller (C). But wait, but if C is T, then C has seen it. But B's statement being false because C has seen it, which is correct. So this case is possible. So another case with one truth-teller.Wait, but let me check again. In case 6:- C=T: C has seen the movie.- A=F: Therefore, A's statement "none have seen it" is false, which is correct because C has.- B=F: His statement is "I have seen it, but both of you haven't." Since B is F, his statement is false. So either he hasn't seen it, or at least one of A or C has. Since C has seen it, that makes B's statement false. So that works.Thus, in case 6, C is T, others F. So one truth-teller.Now, are there any other possible cases where there are two truth-tellers? Let me check again.Wait, initially I considered all cases where at least one T and one F. But maybe I missed some possibilities. Let's see:Wait, when we considered case 1: A=T, B=T, C=F, but that led to contradiction. Similarly, case 4: A=F, B=T, C=T also led to contradiction. So no case with two truth-tellers works. Let me verify.Suppose two truth-tellers:Scenario 1: A=T, B=T, C=F.As before, A says none have seen. If A is T, then none have. B says he has seen it but others haven't. But if B is T, then he has seen it, which contradicts A's statement. Therefore, impossible.Scenario 2: A=T, B=F, C=T.A says none have seen. C says he has seen. If both A and C are T, this is a contradiction. So impossible.Scenario 3: A=F, B=T, C=T.B says he has seen and others haven't. C says he has seen. If both B and C are T, then C has seen it, which contradicts B's statement that the others (including C) haven't. Contradiction.Scenario 4: A=F, B=T, C=F.This is case 5, which is possible but only B is T.Scenario 5: A=F, B=F, C=T.This is case 6, possible with C as T.Scenario 6: A=T, B=F, C=F.Case 3, possible with A as T.So all possible two-truth-teller scenarios lead to contradictions except when the two truth-tellers' statements don't conflict. But in this problem, the statements do conflict. For example, if A and C are both T, A says none have seen, C says he has. Contradiction. Similarly, B and C can't both be T because B says he has seen and others haven't, but C says he has. So only possible cases with one truth-teller: cases 3,5,6.Wait, but wait: Let me check if there's a scenario with two truth-tellers where their statements don't conflict. Suppose A=F, B=T, C=T. But as above, that leads to contradiction. Similarly, others.Alternatively, maybe if two of them are truth-tellers but their statements are compatible. Let's see:Is there any possible way? Let's suppose B and C are both T.If B is T, then B has seen the movie, and A and C have not. But if C is T, then C has seen the movie. Contradiction. So that's impossible.If A and B are both T: A says none have seen. B says he has seen. Contradiction.If A and C are both T: A says none have seen, C says he has. Contradiction.So no, there's no way to have two truth-tellers. Therefore, only possible scenarios are those with one truth-teller: cases 3,5,6.But let's check case 3 again:Case3: A=T, B=F, C=F.A's statement: none have seen. So A, B, C all haven't. Then B's statement is "I have seen, but you two haven't." Since B is F, his statement is false. So either he hasn't seen it, or at least one of A or C has. But since A is T, A hasn't. C is F, but C's statement "I have seen" is false, so C hasn't. Therefore, B's statement is false because he claims to have seen it, but he hasn't. So that works.In this case, all three haven't seen the movie. So A is T, B and C are F. So one truth-teller.Case5: A=F, B=T, C=F.A's statement is false, so at least one has seen. B says he has, and others haven't. Since B is T, he has seen it, and A and C haven't. C is F, so his statement "I have seen" is false, which matches. So yes, B is T, others F. One truth-teller.Case6: A=F, B=F, C=T.A's statement is false, so at least one has seen. B's statement is false. C's statement is true: C has seen it. So in this scenario, C has seen it, A and B haven't. This satisfies all conditions: A=F (so at least one has seen), B=F (his claim to have seen and others not is false because he hasn't and C has), C=T. So one truth-teller.Therefore, all possible valid cases have exactly one truth-teller. Therefore, the answer should be 1.But wait, let me make sure there's no other possibility where there are two truth-tellers that I missed. For example, if there's a case where two are truth-tellers but their statements don't conflict. Let me think.Suppose A=F, B=F, C=T. Wait, that's case6 with one truth-teller.Alternatively, if A=F, B=T, C=T, but that's impossible because B and C would conflict. So no, no two truth-tellers possible.So the only possible valid scenarios are those with exactly one truth-teller. Therefore, the answer must be 1.

✅:To determine how many friends are always right, we analyze each possible combination of truth-tellers (T) and liars (F) under the constraint that at least one is T and one is F. Here's the breakdown:1. Case Analysis: - Case 1 (A=T, B=T, C=F): Contradiction (B claims to have seen the movie, conflicting with A’s statement). - Case 2 (A=T, B=F, C=T): Contradiction (C’s truthful claim to have seen the movie conflicts with A’s statement). - Case 3 (A=T, B=F, C=F): Valid. A truthfully states no one saw the movie. B and C lie, consistent with not having seen it. 1 truth-teller. - Case 4 (A=F, B=T, C=T): Contradiction (B and C’s truthful claims conflict). - Case 5 (A=F, B=T, C=F): Valid. B truthfully claims to have seen the movie, while A and C lie. 1 truth-teller. - Case 6 (A=F, B=F, C=T): Valid. C truthfully claims to have seen the movie, while A and B lie. 1 truth-teller.2. Conclusion: All valid scenarios (Cases 3, 5, 6) result in exactly one truth-teller. No combination with two truth-tellers is possible due to conflicting statements.Answer: boxed{1}

🔑:1. Analysis of Statements: - Let's denote the friends as ( F_1 ), ( F_2 ), and ( F_3 ). - The statements given by each friend are: - ( F_1 ): "None of us have seen this movie." - ( F_2 ): "I have seen this movie, and neither of you have." - ( F_3 ): "I have seen this movie."2. Break Down Each Case: - Each friend either always tells the truth or always lies. - We are informed that at least one friend is always correct and one is always wrong. 3. Case 1: ( F_1 ) Tells the Truth: - If ( F_1 ) is telling the truth, then *none* of the friends have seen the movie. - This means ( F_2 )'s and ( F_3 )'s statements must be false: - False statement by ( F_2 ): It means ( F_2 ) is lying about having seen the movie and lying about others not seeing the movie. - False statement by ( F_3 ): It means ( F_3 ) has not seen the movie. - This is consistent because ( F_1 )'s assertion that nobody has seen the film means ( F_2 ) and ( F_3 ) are lying.4. Conclusion from Case 1: - If ( F_1 ) is always right, then ( F_2 ) and ( F_3 ) are always wrong. This satisfies the constraint of having at least one person tell the truth and one lie.5. Case 2: ( F_2 ) Tells the Truth: - If ( F_2 ) is telling the truth, then: - ( F_2 ) has seen the movie. - ( F_1 ) and ( F_3 ) have not seen the movie. - Analyze ( F_1 )'s and ( F_3 )'s statements given ( F_2 ) is truthful: - ( F_1 )'s statement ("None of us have seen this movie") is false because ( F_2 ) has seen the movie. - ( F_3 )'s statement ("I have seen the movie") is false because ( F_3 ) did not see the movie.6. Conclusion from Case 2: - If ( F_2 ) is always right, then ( F_1 ) and ( F_3 ) are always wrong. This also satisfies the constraint of at least one truth and one lie.7. Case 3: ( F_3 ) Tells the Truth: - If ( F_3 ) is telling the truth, then: - ( F_3 ) has seen the movie. - Analyze statements of ( F_1 ) and ( F_2 ) given ( F_3 ) is truthful: - ( F_1 )'s statement ("None of us have seen this movie") is false because ( F_3 ) has seen it. - ( F_2 )'s statement ("I have seen the movie, and neither of you have") must be completely false: - Because ( F_3 ) saw the movie, the entirety of ( F_2 )'s statement is false.8. Conclusion from Case 3: - If ( F_3 ) is always right, then ( F_1 ) and ( F_2 ) are always wrong. This also satisfies the constraints.9. Conclusion: - In each valid scenario, only one of the friends can be telling the truth, and the other two must be lying. Therefore, the problem confirms that exactly one friend always tells the truth.[boxed{1}]

🧠:Okay, so I have these two problems to solve about finding solutions to quadratic congruences modulo primes of specific forms. Let me start by tackling the first one.Problem (1): Prove that when the prime ( p = 4m + 3 ) and (left(frac{a}{p}right) = 1), then ( x_{0} = pm a^{m+1} ) is a solution to ( x^{2} equiv a pmod{p} ).Alright, I remember that the Legendre symbol (left(frac{a}{p}right) = 1) means that (a) is a quadratic residue modulo (p), so there exists some (x) such that (x^2 equiv a mod p). The prime here is of the form (4m + 3), which is a prime congruent to 3 modulo 4. I think this relates to Euler's criterion, which states that if (p) is an odd prime, then (a^{(p-1)/2} equiv left(frac{a}{p}right) mod p). Since (left(frac{a}{p}right) = 1), Euler's criterion tells us that (a^{(p-1)/2} equiv 1 mod p).Given (p = 4m + 3), then ((p - 1)/2 = (4m + 2)/2 = 2m + 1). So Euler's criterion gives (a^{2m + 1} equiv 1 mod p). Hmm, so that means (a^{2m + 1} equiv 1 mod p). If I multiply both sides by (a), I get (a^{2m + 2} equiv a mod p). Then, (a^{2m + 2} = (a^{m + 1})^2), so ((a^{m + 1})^2 equiv a mod p). Therefore, (x_0 = pm a^{m + 1}) would satisfy (x^2 equiv a mod p). That seems straightforward. Let me check if I did that correctly.Wait, let's step through it again. Since (p = 4m + 3), then ((p - 1)/2 = 2m + 1). Because (a) is a quadratic residue, (a^{2m + 1} equiv 1 mod p). So, multiplying both sides by (a) gives (a^{2m + 2} equiv a mod p). Then (a^{2m + 2}) is (a^{2(m + 1)}), which is ((a^{m + 1})^2). Therefore, ((a^{m + 1})^2 equiv a mod p), so indeed, (x = pm a^{m + 1}) are solutions. That seems right. So the key idea here is using Euler's criterion and manipulating the exponent to get a square.Problem (2): When ( p = 8m + 5 ) and (left(frac{a}{p}right) = 1), find the solution to ( x^{2} equiv a pmod{p} ).Alright, similar idea here, but now the prime is (8m + 5). So let me recall if there's a standard method for primes of the form 8m + 5. Maybe using Euler's criterion again, but perhaps with a different exponent?Given (left(frac{a}{p}right) = 1), Euler's criterion says (a^{(p - 1)/2} equiv 1 mod p). Let's compute ((p - 1)/2) for (p = 8m + 5). That would be ((8m + 4)/2 = 4m + 2). So, (a^{4m + 2} equiv 1 mod p). If I can somehow express the solution (x) in terms of (a) raised to some power, similar to the first problem.In the first problem, we multiplied both sides by (a) to get an exponent that was even, allowing us to take a square root. Maybe here we need a different exponent. Let's think.Suppose we try to raise (a) to some power such that when squared, it gives (a). Let's suppose the solution is (x = pm a^k) for some exponent (k). Then (x^2 = a^{2k} equiv a mod p). So we need (a^{2k} equiv a mod p), which implies (a^{2k - 1} equiv 1 mod p). Since (a^{(4m + 2)} equiv 1 mod p) from Euler's criterion, we can set (2k - 1) to be a multiple of (4m + 2). Let's solve for (k) such that (2k - 1 equiv 0 mod (4m + 2)). That would mean (2k equiv 1 mod (4m + 2)). Let's check if this is possible.Since (4m + 2) is even, and 2 and 4m + 2 are not coprime (they share a factor of 2), the congruence (2k equiv 1 mod (4m + 2)) only has a solution if 2 divides 1, which it doesn't. So that approach might not work. Hmm.Alternatively, maybe we need a different approach. Let's think about the structure of primes congruent to 5 mod 8. I recall that for primes of the form 8m + 5, there is a method to compute square roots using exponentiation. Maybe similar to the case when p ≡ 3 mod 4, but slightly different.Let me recall that when p ≡ 5 mod 8, there's a formula for square roots. Let me check my memory. One method is to note that since p ≡ 5 mod 8, (p - 1)/2 = 4m + 2. Then, perhaps the exponent is (p + 3)/8 or something like that? Wait, let's compute (p + 3)/8. For p = 8m + 5, (8m + 5 + 3)/8 = (8m + 8)/8 = m + 1. Not sure if that helps.Alternatively, let's use the fact that from Euler's criterion, (a^{4m + 2} equiv 1 mod p). Then, (a^{4m + 3} equiv a mod p). So, (a^{4m + 3} equiv a), which implies that (a^{4m + 2} equiv 1). But how does that help?Alternatively, perhaps we can write the exponent such that when squared, it gives a. Let's suppose that x = a^{(4m + 2 + 1)/2}? Wait, 4m + 2 is (p - 1)/2, so (p - 1)/2 + 1 = (p + 1)/2. So, (p + 1)/2 = (8m + 5 + 1)/2 = 4m + 3. So x = a^{4m + 3}? Then x^2 = a^{8m + 6} = a^{8m + 5 + 1} = a^{p} * a. But by Fermat's Little Theorem, a^{p} ≡ a mod p. Therefore, x^2 = a^{p} * a ≡ a * a = a^2 mod p. Wait, that gives x^2 ≡ a^2 mod p, which would mean x ≡ ±a mod p. But that's trivial. So that's not helpful.Hmm. Maybe a different approach. Let's consider that p ≡ 5 mod 8, so 8 divides (p - 5). Let me think about the Tonelli-Shanks algorithm, which is a general method for finding square roots modulo primes. But since the problem is asking for a specific formula, maybe there's a simpler expression.Wait, here's an idea. For primes p ≡ 5 mod 8, there's a known result that one of the square roots of a is either ( a^{(p + 3)/8 } ) or ( a^{(p + 3)/8 } cdot 2^{(p - 1)/4} mod p ). Let me check this.Alternatively, another approach. Let's let k = (p - 5)/8, so p = 8k + 5. Then, (p - 1)/2 = 4k + 2. So Euler's criterion gives a^{4k + 2} ≡ 1 mod p. Therefore, a^{4k + 3} ≡ a mod p. So a^{4k + 3} is a solution? Let's see: (a^{4k + 3})^2 = a^{8k + 6} = a^{8k + 5 + 1} = a^{p} * a. By Fermat's Little Theorem, a^{p} ≡ a mod p, so this becomes a * a = a^2 mod p. So again, (a^{4k + 3})^2 ≡ a^2 mod p, which is not helpful. Hmm.Wait, maybe I need to take a different exponent. Let's consider that (p - 1)/4 = 2m + 1 when p = 8m + 5. Since (p - 1)/4 = (8m + 5 - 1)/4 = (8m + 4)/4 = 2m + 1. So, 2m + 1 is an integer. Let me compute a^{(2m + 1)}. If I square that, I get a^{4m + 2} ≡ 1 mod p. So, a^{2m + 1} ≡ ±1 mod p. But since a is a quadratic residue, maybe a^{2m + 1} ≡ 1? Wait, but from Euler's criterion, we already have a^{4m + 2} ≡ 1, which is (a^{2m + 1})^2 ≡ 1. So that gives a^{2m + 1} ≡ ±1 mod p. If a^{2m + 1} ≡ 1, then maybe we can use that to find the square root. If it's -1, then perhaps we need to adjust.Wait, let's think. Suppose we test ( x = a^{m + 1} ). Then, x^2 = a^{2m + 2}. If I can relate 2m + 2 to the exponent in Euler's criterion. Since p = 8m + 5, so 4m + 2 = (p - 1)/2. So, 2m + 1 = (p - 5)/4. Therefore, a^{(p - 5)/4} ≡ ±1 mod p. Not sure.Alternatively, let me recall that for primes p ≡ 5 mod 8, one method to compute square roots is as follows: Let d = a^m mod p. Then, if d^2 ≡ 1 mod p, then x ≡ ±a^{m + 1} mod p. Otherwise, x ≡ ±2^{2m + 1} a^{m + 1} mod p. Wait, but I need to verify this.Wait, perhaps let me look up the standard formula. Wait, I can't actually look things up, but from my memory, when p ≡ 5 mod 8, the solutions are given by ( x = pm a^{m + 1} ) if ( a^{2m + 1} equiv 1 mod p ), otherwise ( x = pm 2^{2m + 1} a^{m + 1} mod p ). But how does this come about?Alternatively, here's a different approach. Let's assume that x = a^{k} for some exponent k. Then x^2 = a^{2k} ≡ a mod p. So, 2k ≡ 1 mod (order of a modulo p). The order of a divides (p - 1)/2 since a is a quadratic residue. Since p = 8m + 5, the order of a divides 4m + 2. Therefore, 2k ≡ 1 mod (ord(a)), but ord(a) could be a divisor of 4m + 2. However, this might not help directly unless we know ord(a).Alternatively, since we know from Euler's criterion that a^{4m + 2} ≡ 1 mod p, so the order of a divides 4m + 2. Let's write 4m + 2 = 2*(2m + 1). So the order of a is a divisor of 2*(2m + 1). If the order is odd, then 2m + 1 is part of it. Hmm, not sure.Wait, let's try a concrete example to see if I can spot a pattern. Let's take p = 5 (which is 8*0 + 5), and take a quadratic residue. Let's say a = 4. Then, the solutions to x^2 ≡ 4 mod 5 are x ≡ 2 or 3. Now, according to some formula, if x = ±a^{m + 1} where m = 0 here (since p = 5 = 8*0 + 5). Then x = ±4^{1} = ±4 mod 5, which is 4 or 1. But 4^2 is 16 ≡ 1 mod 5, which is not 4. So that doesn't work. Hmm, so maybe that approach isn't correct. Therefore, my previous thought might be wrong.Alternatively, let's take another prime, p = 13 (which is 8*1 + 5). Let's take a quadratic residue, say a = 4. Then solutions to x^2 ≡ 4 mod 13 are x ≡ 2 or 11. Let's compute m = (p - 5)/8 = (13 - 5)/8 = 1. Then according to the formula, maybe x = ±a^{m + 1} = ±4^{2} = ±16 ≡ ±3 mod 13. But 3^2 = 9 ≡ 9 mod 13, which is not 4. So that's not working. So maybe the exponent is different.Alternatively, let's see if there's another exponent. For p = 13, m = 1. Let's try k = (p + 3)/8 = (16)/8 = 2. Then x = a^k = 4^2 = 16 ≡ 3 mod 13. But 3^2 ≡ 9 mod 13. Not helpful. Alternatively, k = (p - 5)/8 + something. Hmm. Alternatively, maybe the exponent is (p + 3)/8. For p = 5, (5 + 3)/8 = 1. So a^1 = 4, which squared is 16 ≡ 1 mod 5. Not 4. Not helpful.Wait, perhaps when p ≡ 5 mod 8, the solutions can be written as x ≡ ±a^{(p + 3)/8} or ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p. Let's test this with p = 5. (p + 3)/8 = 1, 2^{(p - 1)/4} = 2^{1} = 2. So for a = 4, which is 4 mod 5. Then, x ≡ ±4^1 = ±4 mod 5, which gives 4 and 1. 4^2 is 16 ≡ 1, 1^2 ≡ 1. Not helpful. But 2^{(p - 1)/4} = 2. Then ±4 * 2 = ±8 ≡ ±3 mod 5. 3^2 = 9 ≡ 4 mod 5. Ah, there we go. So for p = 5, a = 4, x ≡ ±3 mod 5, which are the solutions. So maybe the formula is x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p.Let's check this formula with another prime. Take p = 13, which is 8*1 + 5. Then (p + 3)/8 = 16/8 = 2, and 2^{(p - 1)/4} = 2^{3} = 8. Let's take a quadratic residue, say a = 4 again. Then x ≡ ±4^2 * 8 mod 13. 4^2 = 16 ≡ 3, 3 * 8 = 24 ≡ 11 mod 13. Then x ≡ ±11 mod 13, which are indeed the solutions (since 11^2 = 121 ≡ 121 - 9*13 = 121 - 117 = 4 mod 13). So that works. Similarly, the other solution would be ±3, but wait, 3 mod 13 is 3, which squared is 9. But 11 is correct. Wait, but 4 has square roots 2 and 11, right? 2 and 11. So according to this formula, ±(a^{(p + 3)/8} * 2^{(p - 1)/4}) gives 11 and 2 (since -11 mod 13 is 2). Wait, 13 - 11 = 2. So yes, 2 is the other solution. So in this case, x ≡ ±11 mod 13, which are 2 and 11. So that works. So the formula seems to hold.Therefore, for primes p ≡ 5 mod 8, if a is a quadratic residue modulo p, then the solutions to x^2 ≡ a mod p are x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p. Alternatively, written in terms of m, since p = 8m + 5, then (p + 3)/8 = (8m + 5 + 3)/8 = m + 1, and (p - 1)/4 = (8m + 5 - 1)/4 = 2m + 1. Therefore, the solution is x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. Alternatively, since 2^{2m + 1} can be written as 2^{(p - 1)/4}.But why does this work? Let me try to derive it.Given that p ≡ 5 mod 8, so p = 8m + 5. Then, (p - 1)/2 = 4m + 2. From Euler's criterion, a^{4m + 2} ≡ 1 mod p. So, a^{4m + 2} ≡ 1 mod p. Let me write this as (a^{2m + 1})^2 ≡ 1 mod p. Therefore, a^{2m + 1} ≡ ±1 mod p.Case 1: If a^{2m + 1} ≡ 1 mod p, then multiplying both sides by a gives a^{2m + 2} ≡ a mod p. Then, (a^{m + 1})^2 ≡ a mod p, so x ≡ ±a^{m + 1} mod p.Case 2: If a^{2m + 1} ≡ -1 mod p, then we need to adjust. Let me recall that 2 is a quadratic residue modulo p if p ≡ ±1 mod 8. Since p ≡ 5 mod 8, 2 is a quadratic non-residue. The Legendre symbol (left(frac{2}{p}right) = -1) when p ≡ 3, 5 mod 8. So 2 is a quadratic non-residue here. Therefore, 2^{(p - 1)/2} ≡ -1 mod p. Since (p - 1)/2 = 4m + 2, so 2^{4m + 2} ≡ -1 mod p. Therefore, (2^{2m + 1})^2 ≡ -1 mod p. So 2^{2m + 1} is a square root of -1 mod p.Therefore, if in Case 2 where a^{2m + 1} ≡ -1 mod p, then multiply both sides by 2^{2m + 1} squared. Wait, let's think. If a^{2m + 1} ≡ -1, then multiply by 2^{4m + 2} ≡ (-1)^2 ≡ 1 mod p. Wait, not sure.Alternatively, since we have a^{2m + 1} ≡ -1, so a^{2m + 1} * 2^{4m + 2} ≡ (-1)(1) ≡ -1 mod p. Not helpful.Wait, let's consider that if a^{2m + 1} ≡ -1 mod p, then (a^{m + 1})^2 = a^{2m + 2} = a * a^{2m + 1} ≡ a * (-1) ≡ -a mod p. So, (a^{m + 1})^2 ≡ -a mod p. So to adjust this, we need to multiply by something that squares to -1. But 2^{2m + 1} squares to -1, as we saw earlier. Therefore, (2^{2m + 1})^2 ≡ -1 mod p. Therefore, if we take x = a^{m + 1} * 2^{2m + 1}, then x^2 = (a^{m + 1})^2 * (2^{2m + 1})^2 ≡ (-a) * (-1) ≡ a mod p.Therefore, in Case 2, x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. So combining both cases, the solutions are either ±a^{m + 1} or ±a^{m + 1} * 2^{2m + 1} mod p, depending on whether a^{2m + 1} is congruent to 1 or -1 mod p.But since the problem states that (left(frac{a}{p}right) = 1), we know that a is a quadratic residue, so x exists. However, depending on a, we might need to use the multiplier 2^{2m + 1}. But how do we know which case we're in? Since the problem is to find a general solution formula without prior knowledge of a, we need a formula that works regardless of the value of a^{2m + 1}. Therefore, the solution can be expressed as x ≡ ±a^{m + 1} * (2^{2m + 1})^{k} mod p, where k is 0 or 1. However, this requires testing whether a^{2m + 1} is 1 or -1. But perhaps there's a deterministic formula.Alternatively, we can note that since 2^{2m + 1} is a square root of -1, and if we take x = a^{m + 1} * (1 + 2^{2m + 1}), but that might not lead us anywhere.Wait, let's recall the formula from earlier. For primes p ≡ 5 mod 8, the square roots of a are given by:x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod pandx ≡ ±a^{(p + 3)/8} mod p.But wait, in our earlier example with p = 13, (p + 3)/8 = 2, and (p - 1)/4 = 3. So 2^{3} = 8. Then a^{2} * 8 mod 13. For a = 4, 4^2 * 8 = 16 * 8 = 128 ≡ 128 - 9*13 = 128 - 117 = 11 mod 13, which is correct. If we take a^{(p + 3)/8} = 4^{2} = 16 ≡ 3 mod 13, which squared is 9, which is not 4. So that case only works when a^{2m + 1} ≡ 1. Therefore, the general solution must account for both possibilities. But how?Alternatively, combining both cases into one formula. For instance, if we let x = a^{m + 1} * 2^{2m + 1 * k}, where k is 0 or 1, depending on the value of a^{2m + 1}. However, since the problem asks for a solution, not necessarily both solutions, but likely a formula similar to the first problem.But maybe there's a different exponent. Let me recall that in the case of p ≡ 5 mod 8, the solution can be written as x ≡ ±a^{(p + 3)/8} mod p if a^{(p - 1)/4} ≡ 1 mod p, and otherwise multiplied by some power of 2. But this requires a case distinction.Alternatively, here's a method inspired by the Tonelli-Shanks algorithm. Let me consider that p - 1 = 8m + 4 = 4*(2m + 1). So, writing p - 1 = Q * 2^s, where Q is odd and s = 2. Here, Q = 2m + 1, s = 2. The Tonelli-Shanks algorithm would then proceed by finding a quadratic non-residue z, which in this case could be 2, since 2 is a quadratic non-residue modulo p. Then, set c = z^Q mod p. But maybe this is getting too involved.Alternatively, let's recall a formula from number theory. For a prime p ≡ 5 mod 8, the solutions to x^2 ≡ a mod p are given by:x ≡ ±a^{(p + 3)/8} mod p, if a^{(p - 1)/4} ≡ 1 mod p,andx ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p, if a^{(p - 1)/4} ≡ -1 mod p.But since (left(frac{a}{p}right) = 1), we know that a^{(p - 1)/2} ≡ 1 mod p. So a^{(p - 1)/4} is a square root of 1 mod p, hence ≡ ±1 mod p. Therefore, depending on which it is, we choose the appropriate formula. However, since the problem just states that a is a quadratic residue, and we need a general solution without knowing a^{(p - 1)/4} in advance, we need a formula that works in either case.But how?Wait, suppose we compute x = a^{(p + 3)/8} mod p. Then x^2 = a^{(p + 3)/4} mod p. But (p + 3)/4 = (8m + 5 + 3)/4 = (8m + 8)/4 = 2m + 2. So x^2 = a^{2m + 2} mod p. But from Euler's criterion, since a^{4m + 2} ≡ 1 mod p, then a^{2m + 2} = a^{2m + 1} * a ≡ (either a or -a) depending on whether a^{2m + 1} is 1 or -1. So x^2 ≡ a^{2m + 2} ≡ a^{2m + 1} * a. If a^{2m + 1} ≡ 1, then x^2 ≡ a, which is what we want. If a^{2m + 1} ≡ -1, then x^2 ≡ -a. To fix this, multiply x by a square root of -1. Since 2^{2m + 1} is a square root of -1 mod p (as shown earlier), then multiplying x by 2^{2m + 1} would give x * 2^{2m + 1}, whose square is (-a)(-1) = a. Therefore, in the case where a^{2m + 1} ≡ -1, x * 2^{2m + 1} is a solution. So, combining both possibilities, the solution can be written as x ≡ ±a^{(p + 3)/8} * (1 or 2^{2m + 1}) mod p. But how to express this without a case distinction?Alternatively, since the possible solutions are either ±a^{(p + 3)/8} or ±a^{(p + 3)/8} * 2^{(p - 1)/4}, and since 2^{(p - 1)/4} ≡ 2^{2m + 1} mod p, which is the square root of -1, we can write the solution as x ≡ ±a^{(p + 3)/8} * (2^{(p - 1)/4})^{k} mod p, where k is 0 or 1. But without knowing k in advance, this isn't a closed-form solution.Alternatively, perhaps the solution is always x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. Wait, but in the case where a^{2m + 1} ≡ 1, this would give x ≡ ±a^{m + 1} * 2^{2m + 1} mod p, but we already know that in that case, x ≡ ±a^{m + 1} mod p. Therefore, this would be incorrect.Alternatively, another idea. Since we know that 2^{2m + 1} is a square root of -1, perhaps we can use this to construct a solution. Suppose we compute x = a^{m + 1} + a^{m + 1} * 2^{2m + 1} mod p. Then, x = a^{m + 1}(1 + 2^{2m + 1}) mod p. Squaring this, we get a^{2m + 2}(1 + 2^{2m + 1})^2. But this seems complicated.Alternatively, let's take a step back. For primes p ≡ 5 mod 8, here's a known formula:If p ≡ 5 mod 8, then the solutions to x² ≡ a mod p are x ≡ ±a^{(p + 3)/8} mod p if a^{(p - 1)/4} ≡ 1 mod p, otherwise x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p.But since we don't know a priori whether a^{(p - 1)/4} is 1 or -1, but since the Legendre symbol is 1, and a^{(p - 1)/2} ≡ 1, then a^{(p - 1)/4} is a square root of 1, so either 1 or -1. Therefore, the solution depends on which it is.However, the problem asks to "find the solution", not necessarily both solutions, but likely a formula that can be computed without prior knowledge. Therefore, perhaps the formula is x ≡ ±a^{(p + 3)/8} mod p if that works, otherwise multiplied by 2^{(p - 1)/4}. But how to combine this into a single expression?Alternatively, here's a trick: compute x ≡ a^{(p + 3)/8} mod p. Then check if x² ≡ a mod p. If yes, done. If not, multiply by 2^{(p - 1)/4} mod p. Since 2^{(p - 1)/4} is a square root of -1, this adjusts the sign.But since the problem likely expects an explicit formula without conditional checks, perhaps the solution is written using both terms. Alternatively, in some references, the solution is given as x ≡ ±a^{m + 1} or ±a^{m + 1} * 2^{2m + 1} mod p. However, since the problem asks to "find the solution", maybe both solutions are expected, but I need to check.Wait, the first problem states the solution as ±a^{m + 1}, so likely the second problem also expects a similar expression. However, the second problem might require a different exponent or multiplier. Let's think again.Wait, going back to the example with p = 5. If m = 0, then the formula in problem (1) would give x = ±a^{1} mod p. But for p = 5, when a = 4, this gives x = ±4 mod 5, which squares to 1 and 16 mod 5, which are 1 and 1. Not correct. Wait, but p = 5 is 4*0 + 5, which isn't of the form 4m + 3. So the first problem is for primes 3 mod 4, the second is 5 mod 8.Wait, but in the first problem, primes are 3 mod 4, and the solution is x = ±a^{m + 1}. For primes 5 mod 8, maybe the solution is x = ±a^{(p + 3)/8} or something else. Let me check another example.Take p = 13 (8*1 + 5), m = 1. Let a = 3. Check if 3 is a quadratic residue modulo 13. 3^{(13-1)/2} = 3^6 = 729 ≡ 729 - 56*13 = 729 - 728 = 1 mod 13. So yes, 3 is a quadratic residue. The solutions to x² ≡ 3 mod 13 are x = ±4 and ±9, since 4² = 16 ≡ 3, 9² = 81 ≡ 3 mod 13.Using the formula: a^{(p + 3)/8} = 3^{(16)/8} = 3^2 = 9 mod 13. So ±9 are solutions. Then 2^{(p - 1)/4} = 2^3 = 8 mod 13. So the other solution would be ±9*8 = ±72 ≡ ±72 - 5*13 = ±72 - 65 = ±7 mod 13. But 7² = 49 ≡ 10 mod 13, which is not 3. Wait, that doesn't work. Hmm, my previous assumption must be wrong.Wait, maybe I made a mistake. For a = 3 in p = 13, the solutions are x = 4 and 9. Let's compute a^{(p + 3)/8} = 3^{2} = 9 mod 13. Then 9 is a solution. Then 9 * 2^{(p - 1)/4} = 9 * 8 = 72 ≡ 72 - 5*13 = 72 - 65 = 7 mod 13. But 7^2 = 49 ≡ 10 mod 13. Not 3. Wait, that's not correct. So perhaps the formula isn't general.Alternatively, maybe I need to use a different non-residue. Wait, 2 is a quadratic non-residue modulo 13. 2^6 = 64 ≡ 12 ≡ -1 mod 13. So 2^{(p - 1)/2} ≡ -1 mod 13. So 2^{(p - 1)/4} = 2^3 = 8 ≡ 8 mod 13. 8^2 = 64 ≡ -1 mod 13. So 8 is a square root of -1.If we take x = a^{(p + 3)/8} * 2^{(p - 1)/4} = 3^2 * 8 = 9 * 8 = 72 ≡ 7 mod 13. Then x^2 = 49 ≡ 10 mod 13. Not helpful. But 4 is the other solution. How is 4 obtained?Wait, let's compute a^{m + 1} = 3^{1 + 1} = 9 mod 13. Then 9 squared is 81 ≡ 3 mod 13. So that works. So in this case, x = ±9 mod 13 are solutions. But the other solution is x = ±4. How is 4 related to 9? 13 - 9 = 4. So -9 ≡ 4 mod 13. So, the solutions are x ≡ ±9 and ±4, but since ±4 are just the negatives of ±9. Wait, no: x = 4 and x = 9 are distinct solutions. Wait, 4^2 = 16 ≡ 3, 9^2 = 81 ≡ 3 mod 13. So both 4 and 9 are solutions. But according to the formula x ≡ ±a^{(p + 3)/8} mod p, we get ±9. But where does 4 come from? It's just the negative of 9 modulo 13? 13 - 9 = 4, so -9 ≡ 4 mod 13. Therefore, the solutions are ±9, which are 4 and 9. So perhaps in this case, the formula works. Because ±9 mod 13 gives 4 and 9. So the formula x ≡ ±a^{(p + 3)/8} mod p gives both solutions. So maybe the answer is x ≡ ±a^{(p + 3)/8} mod p. Let me check this with another example.Take p = 5, a = 4. Then (p + 3)/8 = 1. So x = ±4^1 = ±4 mod 5. 4^2 = 16 ≡ 1 mod 5, which is not 4. Wait, this is not correct. But earlier we saw that 3^2 ≡ 4 mod 5. So the formula fails here. What's the difference?Ah, in this case, a = 4, p = 5. Then a^{(p - 1)/4} = 4^{1} = 4 ≡ -1 mod 5. Therefore, according to the case distinction, we should multiply by 2^{(p - 1)/4} = 2^{1} = 2. So x = ±4^{1} * 2 = ±8 ≡ ±3 mod 5. Which are the correct solutions ±3. But if we use the formula x ≡ ±a^{(p + 3)/8} mod p, we get ±4 mod 5, which are not solutions. Therefore, the formula only works when a^{(p - 1)/4} ≡ 1 mod p. When it's ≡ -1, we need to multiply by 2^{(p - 1)/4}.But since the problem says "find the solution", and doesn't specify that we need a single formula that works in all cases without prior knowledge of a^{(p - 1)/4}, how can we reconcile this?Perhaps the answer is that the solutions are x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4 * k} mod p, where k is 0 or 1 depending on the value of a^{(p - 1)/4}. But since the problem is likely expecting a formula similar to the first problem, which gives a direct expression without case analysis, maybe there's a different approach.Wait, I found a resource once that mentioned for p ≡ 5 mod 8, the solutions are x ≡ ±a^{m + 1} mod p if a^{2m + 1} ≡ 1 mod p, else x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. But since the problem doesn't specify conditions on a beyond being a quadratic residue, perhaps the solution requires both possibilities. But the question says "find the solution", not "find all solutions". Since there are two solutions, maybe the answer is expressed in terms of both.Alternatively, here's another approach. Let's note that for primes p ≡ 5 mod 8, since 8 divides (p - 5), we can write p = 8m + 5. Then, (p - 1)/2 = 4m + 2. Let's set x = a^{k} for some exponent k. Then, x^2 = a^{2k} ≡ a mod p. Therefore, we need a^{2k - 1} ≡ 1 mod p. From Euler's criterion, we know a^{4m + 2} ≡ 1 mod p. Therefore, 2k - 1 must be a multiple of 4m + 2. So, 2k - 1 = t*(4m + 2), for some integer t. Solving for k: 2k = 1 + t*(4m + 2). Therefore, k = (1 + t*(4m + 2))/2. We need k to be an integer. Since 4m + 2 is even, t*(4m + 2) is even, so 1 + even is odd, so k would be a half-integer. Which is not possible. Therefore, there is no integer k such that a^{2k - 1} ≡ 1 mod p. Therefore, this approach doesn't work.Alternatively, perhaps we can find k such that 2k ≡ 1 mod (ord(a)), where ord(a) is the order of a modulo p. Since ord(a) divides 4m + 2, which is even, and since a is a quadratic residue, ord(a) divides (p - 1)/2 = 4m + 2. So, 2k ≡ 1 mod ord(a). Since ord(a) is even (because 4m + 2 is even), and 2 and ord(a) share a common factor of 2, this equation has a solution only if ord(a) divides 1, which is impossible. Hence, there is no solution for k in this case. Therefore, my initial approach is flawed.Perhaps I need to abandon the exponent approach and think differently. Let's consider that for primes p ≡ 5 mod 8, there is a method to find square roots using the following identity:If p ≡ 5 mod 8, then one solution to x² ≡ a mod p is given by:x ≡ a^{(p + 3)/8} mod p, if a^{(p - 1)/4} ≡ 1 mod p,andx ≡ a^{(p + 3)/8} * 2^{(p - 1)/4} mod p, if a^{(p - 1)/4} ≡ -1 mod p.But since we don't know a^{(p - 1)/4} in advance, but we can compute it. However, the problem likely expects an answer that can be expressed without conditional checks. Therefore, perhaps the solution is presented as x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4 * k} mod p, where k is 0 or 1. But this is not a precise formula.Alternatively, recall that in the first problem, the exponent was (m + 1), which is ( (p - 3)/4 + 1 ) = (p + 1)/4. For the second problem, maybe the exponent is related to (p + 3)/8.Let me compute (p + 3)/8 for p = 8m + 5: (8m + 5 + 3)/8 = m + 1. So, a^{(m + 1)}. Then, squaring this gives a^{2m + 2}. From Euler's criterion, a^{4m + 2} ≡ 1 mod p. So, a^{2m + 2} = a^{2m + 1} * a. Therefore, a^{2m + 2} ≡ a or -a mod p, depending on a^{2m + 1}. Therefore, if we let x = a^{m + 1} * c, where c is either 1 or 2^{2m + 1}, then x^2 ≡ a or (-a)*(-1) = a. Therefore, the solution is x ≡ ±a^{m + 1} * (1 or 2^{2m + 1}) mod p. Since 2^{2m + 1} is a square root of -1, multiplying by it effectively corrects the sign.But how to write this without conditioning? Perhaps the solution is x ≡ ±a^{m + 1} * 2^{(p - 5)/8 * 4} mod p. Wait, but (p - 5)/8 = m, so 4m. But 2^{4m} might not relate. Alternatively, since 2^{2m + 1} is the square root of -1, and if we take x = a^{m + 1} * 2^{2m + 1}, then x^2 = a^{2m + 2} * (-1) = (-a^{2m + 2}). But unless a^{2m + 2} ≡ -a, which would make x² ≡ a. Wait, this is getting too convoluted.Let me try to summarize. For primes p ≡ 5 mod 8, the solutions to x² ≡ a mod p are given by:x ≡ ±a^{(p + 3)/8} mod p if a^{(p - 1)/4} ≡ 1 mod p,andx ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p otherwise.Since the problem states that a is a quadratic residue, we know that a^{(p - 1)/2} ≡ 1 mod p, so a^{(p - 1)/4} ≡ ±1 mod p. Therefore, the solution depends on this value, but without knowing it in advance, we can't choose which formula to use. However, the problem asks to "find the solution", which likely expects a formula that can be computed directly. Given that, perhaps the standard solution is expressed as x ≡ ±a^{m + 1} * 2^{2m + 1} mod p, but in our earlier example with p = 13 and a = 3, this would be 3^{2} * 2^{3} = 9 * 8 = 72 ≡ 7 mod 13, which is not a solution. So that's incorrect.Wait, but in the first problem, the exponent was m + 1, which for p = 4m + 3 is (p - 3)/4 + 1 = (p + 1)/4. For the second problem, p = 8m + 5, so m here is different. Let's denote m in the second problem as n, so p = 8n + 5. Then, m = (p - 5)/8. Then, (p + 3)/8 = n + 1. So, a^{(n + 1)}. Then, squaring gives a^{2n + 2}. Let's check with p = 13, which would be n = 1. So, a^{2} and 2^{3} = 8. For a = 3, 3^2 * 8 = 9 * 8 = 72 ≡ 7 mod 13, which squared is 49 ≡ 10 mod 13. Not correct. So that's not working.Alternatively, perhaps x = a^{2m + 1}. For p = 8m + 5, then x = a^{2m + 1}. Then, x^2 = a^{4m + 2} ≡ 1 mod p. Which gives x^2 ≡ 1, so x ≡ ±1 mod p. Not helpful.Alternatively, x = a^{m + 1} + a^{m + 1}. Doesn't make sense.Alternatively, let's revisit the initial example with p = 5, a = 4. The solutions are x ≡ ±3 mod 5. According to the formula x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p, we have (5 + 3)/8 = 1, 2^{(5 - 1)/4} = 2^1 = 2. So x ≡ ±4^1 * 2 = ±8 ≡ ±3 mod 5. That works. For p = 13, a = 4: (13 + 3)/8 = 2, 2^{(13 - 1)/4} = 2^3 = 8. So x ≡ ±4^2 * 8 = ±16 * 8 = ±128 ≡ ±128 - 9*13 = ±128 - 117 = ±11 mod 13. Which is correct. For a = 3 in p = 13: x ≡ ±3^2 * 8 = ±9 * 8 = ±72 ≡ ±7 mod 13. But 7^2 = 49 ≡ 10 mod 13, which is not 3. Wait, but 3's square roots are 4 and 9. So this formula doesn't yield those. What's the problem here?Ah, because in this case, a^{(p - 1)/4} ≡ 1 mod p. Let's compute a^{(p - 1)/4} = 3^{(13 - 1)/4} = 3^{3} = 27 ≡ 1 mod 13. Therefore, according to the case distinction, we should use x ≡ ±a^{(p + 3)/8} = ±3^{2} = ±9 mod 13, which gives 9 and 4 (since -9 ≡ 4 mod 13). So in this case, the formula x ≡ ±a^{(p + 3)/8} works, and multiplying by 2^{(p - 1)/4} gives the wrong solution. Therefore, the correct formula depends on a^{(p - 1)/4}. But how to write this without knowing it in advance?Perhaps the answer is that the solutions are x ≡ ±a^{(p + 3)/8} mod p if a^{(p - 1)/4} ≡ 1 mod p, and x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p otherwise. But since the problem asks to "find the solution", it's expecting a general formula. Given that the first problem had a straightforward formula, maybe there's a similar one here.Wait, here's another approach. Note that since p ≡ 5 mod 8, then 2 is a quadratic non-residue, and -1 is a quadratic residue if and only if p ≡ 1 mod 4. Since p ≡ 5 mod 8, p ≡ 1 mod 4, so -1 is a quadratic residue. However, 2 is a quadratic non-residue.Let me try to construct a solution using the identity:If we let x = a^{(p + 3)/8} * (1 + 2^{(p - 1)/4}) mod p, then x^2 = a^{(p + 3)/4} * (1 + 2^{(p - 1)/4})^2. But this seems complicated.Alternatively, since 2^{(p - 1)/4} ≡ sqrt(-1) mod p, then perhaps x = a^{(p + 3)/8} * (1 + sqrt(-1)) mod p. But we don't have sqrt(-1) unless we compute it, which circles back.Alternatively, let's take inspiration from the first problem. For p ≡ 3 mod 4, the solution is x ≡ ±a^{(p + 1)/4} mod p. For p ≡ 5 mod 8, which is 5 mod 4, but since 5 ≡ 1 mod 4, perhaps a different approach is needed. Wait, p ≡ 5 mod 8 implies p ≡ 1 mod 4. So maybe there's a formula for primes p ≡ 1 mod 4, but that's more general.However, primes of the form 8m + 5 are a subset of primes ≡ 1 mod 4. The general solution for primes ≡ 1 mod 4 involves more complex algorithms like Tonelli-Shanks or Cipolla's algorithm. But since the problem specifies p ≡ 5 mod 8, there must be a simpler formula.After some research in my memory, I recall that for primes p ≡ 5 mod 8, the solutions can be expressed as x ≡ ±a^{m + 1} mod p if a^{2m + 1} ≡ 1 mod p, otherwise x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. But since this requires knowing the value of a^{2m + 1}, which is not provided, the answer must accommodate this by including the term with 2^{2m + 1}. However, the problem states that (left(frac{a}{p}right) = 1), but doesn't specify further conditions on a. Therefore, the general solution must account for both possibilities.But how to present this? Since the problem asks to "find the solution", and in the first part, it's ±a^{m + 1}, maybe in this case, the solution is similar but includes a factor involving powers of 2. Based on the examples and derivations, the correct formula when a^{(p - 1)/4} ≡ -1 is to multiply by 2^{(p - 1)/4}. Therefore, the solution can be written as:x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4 * k} mod p, where k = 0 or 1.But since the problem likely expects a specific expression, perhaps combining both cases, it might be written as x ≡ ±a^{m + 1} or ±a^{m + 1} * 2^{2m + 1} mod p. However, in terms of p, substituting m = (p - 5)/8, the exponents become:a^{m + 1} = a^{(p - 5)/8 + 1} = a^{(p + 3)/8},and2^{2m + 1} = 2^{(2(p - 5)/8 + 1)} = 2^{(p - 5)/4 + 1} = 2^{(p - 5 + 4)/4} = 2^{(p - 1)/4}.Therefore, the solutions are x ≡ ±a^{(p + 3)/8} mod p or x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p.Thus, the answer is x ≡ ±a^{(p + 3)/8} or ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p. However, since the problem asks for "the solution", not "all solutions", but there are two solutions, perhaps the answer expects both possibilities expressed with ±.Alternatively, since the problem might expect the solution expressed as x ≡ ±a^{(p + 3)/8} mod p if a^{(p - 1)/4} ≡ 1, and otherwise multiplied by 2^{(p - 1)/4}, but since the user asked to find the solution given that a is a quadratic residue, the answer is likely the general formula incorporating the necessary terms.Therefore, combining the cases, the solution is x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4 * k} mod p, where k is 0 or 1. But since this requires testing whether a^{(p - 1)/4} is 1 or -1, and the problem doesn't provide further information, the answer must be presented as two possible solutions.However, in the first problem, the solution is given as ±a^{m + 1}, which works universally for primes 3 mod 4. For primes 5 mod 8, it's conditional, but since the problem asks to "find the solution", perhaps the answer is expected to mirror the first problem's structure but adjusted for the different prime form.Given the examples and derivations above, the correct formula when p ≡ 5 mod 8 is:x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p.But wait, in the case when a^{(p - 1)/4} ≡ 1, this formula would give x ≡ ±a^{(p + 3)/8} * 1 mod p, which is correct. When a^{(p - 1)/4} ≡ -1, then x ≡ ±a^{(p + 3)/8} * (-1) mod p, but this isn't helpful. Wait, no. 2^{(p - 1)/4} is sqrt(-1), not (-1). Therefore, multiplying by 2^{(p - 1)/4} gives the necessary adjustment.Actually, let's re-express the solution in terms of p. Let me compute (p + 3)/8 and (p - 1)/4.Given p = 8m + 5,(p + 3)/8 = (8m + 5 + 3)/8 = m + 1,(p - 1)/4 = (8m + 5 - 1)/4 = 2m + 1.Therefore, the solution can be written as:x ≡ ±a^{m + 1} * 2^{2m + 1} mod p.But in the case when a^{2m + 1} ≡ 1, then x ≡ ±a^{m + 1} mod p suffices. However, when a^{2m + 1} ≡ -1, then x ≡ ±a^{m + 1} * 2^{2m + 1} mod p. Since 2^{2m + 1} squared is -1, this multiplication corrects the sign.But since the problem asks to find the solution given that a is a quadratic residue, without further conditions, the answer must account for both possibilities. However, combining these into a single formula requires a case distinction, which isn't elegant.After careful consideration and examples, I think the correct answer that incorporates both cases is:x ≡ ±a^{m + 1} or x ≡ ±a^{m + 1} * 2^{2m + 1} mod p.But to express this in terms of p, substituting m = (p - 5)/8,x ≡ ±a^{(p + 3)/8} or x ≡ ±a^{(p + 3)/8} * 2^{(p - 1)/4} mod p.Therefore, the solutions are given by these two cases. Since the problem asks to "find the solution", and there are two solutions, this is the most accurate answer.

✅:Problem (1) Solution:Given a prime ( p = 4m + 3 ) where ( left(frac{a}{p}right) = 1 ), we use Euler's criterion: ( a^{(p-1)/2} equiv 1 pmod{p} ). Substituting ( (p-1)/2 = 2m + 1 ), we have ( a^{2m+1} equiv 1 pmod{p} ). Multiplying both sides by ( a ): ( a^{2m+2} equiv a pmod{p} ). This simplifies to ( (a^{m+1})^2 equiv a pmod{p} ). Thus, ( x_0 = pm a^{m+1} ) are solutions.Final Answer (1): ( boxed{x equiv pm a^{m+1} pmod{p}} )---Problem (2) Solution:For prime ( p = 8m + 5 ) with ( left(frac{a}{p}right) = 1 ), Euler's criterion gives ( a^{4m+2} equiv 1 pmod{p} ). Let ( k = (p+3)/8 = m+1 ). Compute ( a^{k} pmod{p} ): - If ( a^{2m+1} equiv 1 pmod{p} ), then ( x = pm a^{k} ). - If ( a^{2m+1} equiv -1 pmod{p} ), multiply by ( 2^{(p-1)/4} = 2^{2m+1} ), a root of (-1), yielding ( x = pm a^{k} cdot 2^{2m+1} ).Thus, the solutions are: ( x equiv pm a^{(p+3)/8} ) or ( x equiv pm a^{(p+3)/8} cdot 2^{(p-1)/4} pmod{p} ).Final Answer (2): ( boxed{x equiv pm a^{frac{p+3}{8}} cdot 2^{frac{p-1}{4}} pmod{p}} )

🔑:# Part (1):Let's analyze the problem where the prime ( p ) is given in the form ( p = 4m + 3 ) and it is known that ( left( frac{a}{p} right) = 1 ). We need to show that ( x_0 = pm a^{m+1} ) is a solution to the congruence ( x^2 equiv a pmod{p} ).1. Consider ( p ) of the form ( p = 4m + 3 ). This implies: [ 4m + 3 equiv 3 pmod{4} ]2. Given that ( left( frac{a}{p} right) = 1 ), it follows that ( a ) is a quadratic residue modulo ( p ). Therefore, there exists an integer ( x_0 ) such that: [ x_0^2 equiv a pmod{p} ]3. Let ( x_0 = a^{m+1} ). We need to verify if ( x_0 ) satisfies the congruence ( x_0^2 equiv a pmod{p} ): [ x_0^2 = (a^{m+1})^2 = a^{2(m+1)} ]4. Simplify the power exponent: [ a^{2(m+1)} = a^{2m + 2} ]5. Since ( p-1 = 4m + 2 ), it follows: [ 2m + 2 = p-1 ]6. Thus ( a^{p-1} equiv 1 pmod{p} ) by Fermat's Little Theorem: [ a^{p-1} equiv 1 pmod{p} ]Thus, we conclude ( x_0 = a^{m+1} ) and similarly ( -a^{m+1} ) (by symmetry in quadratic residues) are solutions. Therefore:[boxed{x_0 = pm a^{m+1}}]# Part (2):For the situation where the prime ( p ) is given in the form ( p = 8m + 5 ) and ( left( frac{a}{p} right) = 1 ), we need to find the solutions to the congruence ( x^2 equiv a pmod{p} ).1. Consider ( p ) of the form ( p = 8m + 5 ). Thus: [ 8m + 5 equiv 5 pmod{8} ]2. Given ( left( frac{a}{p} right) = 1 ), ( a ) is a quadratic residue modulo ( p ).3. Depending on the case: i) Case 1: ( a^{2m+1} equiv 1 pmod{p} ): [ x_0 equiv pm a^{m+1} pmod{p} ] ii) Case 2: ( a^{2m+1} equiv -1 pmod{p} ): [ x_0 equiv pm 2^{2m+1} a^{m+1} pmod{p} ]4. To prove that ( left( frac{-2}{p} right) = -1 ), consider the Legendre symbol which uses properties of quadratic residues: [ left( frac{-2}{p} right) = left( frac{-1}{p} right) left( frac{2}{p} right) ]By the properties and specific calculations using Gauss's Lemma: [ left( frac{-1}{p} right) = (-1)^{frac{p-1}{2}} = (-1)^{4m+2} = (-1)^2 = 1 ] [ left( frac{2}{p} right) = (-1)^{frac{p^2-1}{8}} = (-1)^{frac{(8m+5)^2-1}{8}} = (-1)^{8m^2+10m+3} = (-1)^{8m(m+1)+3} = (-1)^{3} = -1 ]Therefore: [ left( frac{-2}{p} right) = left( frac{-1}{p} right) left( frac{2}{p} right) = 1 cdot (-1) = -1 ]So, the solution form varies based on calculating ( a^{2m+1} ) modulo ( p ) and splits into two cases as described above.Thus, the needed values for ( x ) satisfying the congruence are:[boxed{x_0 = pm a^{m+1}} quad text{or} quad boxed{x_0 equiv pm 2^{2m+1} a^{m+1} pmod{p}}]

🧠:Okay, let's see. The problem is asking for the possible values of base ( b ) such that the number ( 121_b ) is a perfect square. The options given are from (A) to (E), with (D) being ( b > 2 ). Hmm. I need to figure out which bases make ( 121_b ) a perfect square. First, I should recall how numbers are represented in different bases. The number ( 121_b ) in base ( b ) translates to what in decimal? Let me think. In base ( b ), the digits represent powers of ( b ). So, the rightmost digit is ( b^0 ), the middle is ( b^1 ), and the leftmost is ( b^2 ). Therefore, ( 121_b ) would be:[1 times b^2 + 2 times b + 1 times 1]Simplifying that, it's:[b^2 + 2b + 1]Wait, that looks familiar. ( b^2 + 2b + 1 ) is a quadratic expression. Let me factor that. Oh, right! It factors into:[(b + 1)^2]So, ( 121_b = (b + 1)^2 ) in decimal. Therefore, no matter what base ( b ) we're in, as long as the digits are valid, the number ( 121_b ) will always be a perfect square because it's equal to ( (b + 1)^2 ). But wait, there must be some constraints on the base ( b ). In any base ( b ), the digits used must be less than the base. So, looking at the number ( 121_b ), the digits are 1, 2, and 1. The digit 2 must be valid in base ( b ), which means the base has to be greater than 2. Because in base 2, the digits can only be 0 and 1, right? So, the digit 2 is not allowed in base 2. Therefore, the base ( b ) must satisfy ( b > 2 ). So, that would mean that for any base ( b ) where ( b > 2 ), the number ( 121_b ) is a perfect square, since ( (b + 1)^2 ) is a square, and the digits are all valid (since 2 is less than ( b )). Looking back at the options given: (D) is ( b > 2 ), which seems to match. Let me double-check the other options to make sure.Option (A) says only base 10. But if, for example, we take base 3. Then ( 121_3 ) is ( 1*9 + 2*3 + 1 = 16 ), which is 4 squared. So, that's a perfect square. Similarly, base 4: ( 1*16 + 2*4 + 1 = 25 = 5^2 ). Also a square. So, (A) is incorrect.Option (B) says only 10 and 5. Let's check base 5. ( 121_5 = 1*25 + 2*5 + 1 = 36 = 6^2 ). Correct. But as shown earlier, base 3 and 4 also work, so (B) is not correct either.Option (C) says ( 2 leq b leq 10 ). But base 2 is invalid because of the digit 2. So, the lower limit is actually ( b > 2 ). Therefore, (C) is incorrect.Option (E) states there is no such ( b ), which is clearly not true because we have examples where it works.Therefore, the correct answer should be (D) ( b > 2 ). Wait, but let me confirm with another example. Take base 16. ( 121_{16} = 1*256 + 2*16 + 1 = 256 + 32 + 1 = 289 = 17^2 ). Yep, that's a square. So even in base 16, which is higher than 10, it's still a perfect square. Therefore, as long as the base is greater than 2, it works. So, answer (D) is correct.I think that's all. All the other options are too restrictive or incorrect. So, the possible values of ( b ) are all bases greater than 2.

✅:The number (121_b) in base (b) converts to the decimal value (1 cdot b^2 + 2 cdot b + 1 = (b + 1)^2), which is inherently a perfect square. However, the digits in the number must be valid for the base. The digit (2) requires the base (b) to satisfy (b > 2). Analysis of options:- (A) Incorrect. Bases like (3, 4, 5), and higher work (e.g., (121_3 = 16 = 4^2)).- (B) Incorrect. Valid bases are not limited to (5) and (10); all (b > 2) are valid.- (C) Incorrect. Base (2) is invalid due to the digit (2).- (D) Correct. Any base (b > 2) ensures valid digits and ((b + 1)^2) is a perfect square.- (E) Incorrect. Counterexamples exist (e.g., (b = 3) yields (16), a perfect square).Final Answer: (boxed{D})

🔑:To solve this problem, let's break down the given information and proceed step-by-step.[121_{b}]is a number in base ( b ). By definition, the value of this number in decimal (base 10) can be expressed as:[ 1 cdot b^2 + 2 cdot b + 1]The problem states that (121_{b}) is equivalent to the square of a whole number, ( k^2 ). Hence, we can write:[1 cdot b^2 + 2 cdot b + 1 = (b+1)^2]Now, let's verify whether ( (b+1)^2 ) actually expands to ( 1 cdot b^2 + 2 cdot b + 1 ):[(b+1)^2 = b^2 + 2b + 1]Indeed, the expanded form of ((b+1)^2) matches the given form ( 1 cdot b^2 + 2 cdot b + 1 ). Step-by-Step Verification:1. Expression in Decimal Form:[ 1 cdot b^2 + 2 cdot b + 1 ]2. Expansion of ((b+1)^2):[(b+1)^2 = b^2 + 2b + 1]This verifies that the given base representation ( 121_{b} ) is indeed ((b+1)^2).3. Determine Possible Values of ( b ): - In a base ( b ) system, the digits must be from 0 to ( b-1 ). - For ( 121_{b} ): - The highest digit used is 2, which implies: - (2 < b)Thus, ( b ) must be greater than 2. Therefore, the possible values for ( b ) are ( b > 2 ).Consequently, the correct conclusion is:[boxed{(D) , b > 2}]