🧠:Okay, so I need to prove that 1986 divides infinitely many terms of the sequence defined by the recurrence relation ( u_{n+2} = u_{n+1}^2 - u_n ), with initial terms ( u_1 = 39 ) and ( u_2 = 45 ). Hmm, 1986 seems like a big number. Let me start by factoring it to see if that helps. First, factoring 1986. Let's divide by 2 first: 1986 ÷ 2 = 993. Then 993: check divisibility by 3. 9 + 9 + 3 = 21, which is divisible by 3. So 993 ÷ 3 = 331. Now, 331 is a prime number, right? Let me check: 331 divided by 2 is not, 3? 3*110=330, so 331 is prime. So 1986 factors into 2 × 3 × 331. Therefore, to show that 1986 divides infinitely many terms of the sequence, it's sufficient to show that each prime power factor (2, 3, and 331) divides infinitely many terms. By the Chinese Remainder Theorem, if there are infinitely many terms divisible by each of these primes, and since they are pairwise coprime, then there are infinitely many terms divisible by their product, which is 1986. So I need to handle each prime factor separately. Let me start with modulo 2. Let's compute the sequence modulo 2. The recurrence is ( u_{n+2} equiv u_{n+1}^2 - u_n mod 2 ). Given u1 = 39 mod 2 is 1, u2 = 45 mod 2 is 1. So:u1 ≡ 1 mod 2u2 ≡ 1 mod 2Compute u3: u2^2 - u1 ≡ 1 - 1 = 0 mod 2u3 ≡ 0 mod 2u4: u3^2 - u2 ≡ 0 - 1 = -1 ≡ 1 mod 2u5: u4^2 - u3 ≡ 1 - 0 = 1 mod 2u6: u5^2 - u4 ≡ 1 - 1 = 0 mod 2u7: u6^2 - u5 ≡ 0 - 1 = -1 ≡ 1 mod 2u8: u7^2 - u6 ≡ 1 - 0 = 1 mod 2u9: u8^2 - u7 ≡ 1 - 1 = 0 mod 2So the pattern modulo 2 is 1, 1, 0, 1, 1, 0, 1, 1, 0, ... repeating every 3 terms. Therefore, every third term starting from u3 is divisible by 2. So 2 divides infinitely many terms.Now modulo 3. Let's compute the sequence modulo 3.u1 = 39 mod 3 is 0, u2 = 45 mod 3 is 0. So:u1 ≡ 0 mod 3u2 ≡ 0 mod 3Compute u3: u2^2 - u1 ≡ 0 - 0 = 0 mod 3Similarly, u4: u3^2 - u2 ≡ 0 - 0 = 0 mod 3So all terms from u1 onwards are 0 mod 3. Wait, is that possible? Let me check:If u1 ≡ 0 mod 3, u2 ≡ 0 mod 3, then u3 = 0^2 - 0 = 0 mod 3. Then u4 = 0^2 - 0 = 0 mod 3, etc. So the entire sequence is 0 mod 3. Therefore, every term is divisible by 3. Hence, infinitely many terms divisible by 3.Now the harder part is 331. Let's check modulo 331. Since 331 is prime. So we need to show that 331 divides infinitely many terms of the sequence.To handle modulo 331, perhaps we can consider the sequence modulo 331 and show that it's periodic modulo 331, and then find a term that is 0 mod 331. Since the sequence is defined by a recurrence relation, modulo a prime, the number of possible pairs (u_n, u_{n+1}) is finite (specifically, 331^2 possibilities). Therefore, by the Pigeonhole Principle, eventually, the pair (u_n, u_{n+1}) must repeat, leading to a periodic sequence. Once it's periodic, if 0 appears once, it will appear infinitely often.So the plan is: Show that in the sequence modulo 331, 0 appears at least once. Then, because the sequence becomes periodic, 0 will appear infinitely often.Therefore, first, compute the sequence modulo 331 until we find a term that is 0 mod 331. But manually computing this might be tedious. Maybe there is a smarter way.Alternatively, perhaps we can look for a cycle where 0 occurs. Alternatively, consider solving the recurrence relation modulo 331.Given the recurrence relation ( u_{n+2} = u_{n+1}^2 - u_n mod 331 ), with initial terms u1 = 39 mod 331 = 39, u2 = 45 mod 331 = 45.We can compute terms step by step modulo 331 until we find a 0 or until the sequence repeats.But doing this manually for 331 terms would be time-consuming, but maybe there's a pattern or a way to find a period.Alternatively, perhaps we can use the fact that in a recurrence relation modulo a prime, the period is at most p^2 - 1. But since 331 is prime, the maximum period is 331^2 - 1. However, that's still a very large number. Alternatively, perhaps we can find a term where u_{k} ≡ 0 mod 331. Let's try to compute the first few terms modulo 331.Given u1 = 39, u2 = 45.Compute u3 = u2^2 - u1 = 45^2 - 39 = 2025 - 39 = 1986. 1986 mod 331: Let's compute 331*6 = 1986. So 1986 mod 331 = 0. So u3 ≡ 0 mod 331.Oh! That's convenient. So u3 is divisible by 331. Then, let's compute u4: u3^2 - u2 ≡ 0^2 - 45 ≡ -45 ≡ 331 - 45 = 286 mod 331.u4 ≡ 286 mod 331.u5 = u4^2 - u3 ≡ 286^2 - 0 ≡ 286^2 mod 331.Compute 286 mod 331 is 286. 286^2: Let's compute this. Let's note that 286 = -45 mod 331 (since 331 - 45 = 286). So (-45)^2 = 2025. 2025 mod 331: 331*6 = 1986, 2025 - 1986 = 39. So 2025 ≡ 39 mod 331. Therefore, u5 ≡ 39 mod 331.u5 ≡ 39 mod 331.Then u6 = u5^2 - u4 ≡ 39^2 - 286. 39^2 = 1521. 1521 mod 331: 331*4 = 1324, 1521 - 1324 = 197. So 1521 ≡ 197 mod 331. Then 197 - 286 = -89 ≡ 331 - 89 = 242 mod 331. So u6 ≡ 242 mod 331.u6 ≡ 242.u7 = u6^2 - u5 ≡ 242^2 - 39. Compute 242 mod 331 is 242. 242^2: Let's compute this. Let's note that 242 = -89 mod 331. So (-89)^2 = 7921. 7921 divided by 331: Let's compute 331*23 = 7613, 7921 - 7613 = 308. So 7921 ≡ 308 mod 331. Therefore, 242^2 ≡ 308 mod 331. Then, 308 - 39 = 269 mod 331. So u7 ≡ 269.u7 ≡ 269 mod 331.u8 = u7^2 - u6 ≡ 269^2 - 242. 269 mod 331 is 269. 269^2: Let's compute. 269^2 = (270 - 1)^2 = 270^2 - 2*270 +1 = 72900 - 540 +1 = 72361. Now divide by 331: 331*218 = 331*200 + 331*18 = 66200 + 5958 = 72158. Then 72361 - 72158 = 203. So 269^2 ≡ 203 mod 331. Then 203 - 242 = -39 ≡ 331 - 39 = 292 mod 331. So u8 ≡ 292.u8 ≡ 292 mod 331.u9 = u8^2 - u7 ≡ 292^2 - 269. 292 mod 331 is 292. 292^2: Let's compute. 292 = -39 mod 331. So (-39)^2 = 1521. As before, 1521 mod 331 is 197. Then 197 - 269 = -72 ≡ 331 - 72 = 259 mod 331. So u9 ≡ 259.u9 ≡ 259 mod 331.u10 = u9^2 - u8 ≡ 259^2 - 292. 259 mod 331 is 259. 259^2: Let's compute. 259^2. Let's note that 259 = -72 mod 331. So (-72)^2 = 5184. 5184 divided by 331: 331*15 = 4965, 5184 - 4965 = 219. So 259^2 ≡ 219 mod 331. Then 219 - 292 = -73 ≡ 331 - 73 = 258 mod 331. So u10 ≡ 258 mod 331.u10 ≡ 258.u11 = u10^2 - u9 ≡ 258^2 - 259. 258 mod 331 is 258. 258^2: 258 = -73 mod 331. (-73)^2 = 5329. 5329 divided by 331: 331*16 = 5296, 5329 - 5296 = 33. So 258^2 ≡ 33 mod 331. Then 33 - 259 = -226 ≡ 331 - 226 = 105 mod 331. u11 ≡ 105.u11 ≡ 105 mod 331.u12 = u11^2 - u10 ≡ 105^2 - 258. 105^2 = 11025. 11025 divided by 331: Let's compute 331*33 = 10923, 11025 - 10923 = 102. So 105^2 ≡ 102 mod 331. 102 - 258 = -156 ≡ 331 - 156 = 175 mod 331. u12 ≡ 175.u12 ≡ 175 mod 331.u13 = u12^2 - u11 ≡ 175^2 - 105. 175^2 = 30625. 30625 divided by 331: Let's see, 331*92 = 331*90 + 331*2 = 29790 + 662 = 30452. 30625 - 30452 = 173. So 175^2 ≡ 173 mod 331. 173 - 105 = 68 mod 331. u13 ≡ 68.u13 ≡ 68 mod 331.u14 = u13^2 - u12 ≡ 68^2 - 175. 68^2 = 4624. 4624 divided by 331: 331*13 = 4303, 4624 - 4303 = 321. So 68^2 ≡ 321 mod 331. 321 - 175 = 146 mod 331. u14 ≡ 146.u14 ≡ 146 mod 331.u15 = u14^2 - u13 ≡ 146^2 - 68. 146 mod 331 is 146. 146^2 = 21316. 21316 divided by 331: Let's compute 331*64 = 21184, 21316 - 21184 = 132. So 146^2 ≡ 132 mod 331. 132 - 68 = 64 mod 331. u15 ≡ 64.u15 ≡ 64 mod 331.u16 = u15^2 - u14 ≡ 64^2 - 146. 64^2 = 4096. 4096 divided by 331: 331*12 = 3972, 4096 - 3972 = 124. So 64^2 ≡ 124 mod 331. 124 - 146 = -22 ≡ 331 - 22 = 309 mod 331. u16 ≡ 309.u16 ≡ 309 mod 331.u17 = u16^2 - u15 ≡ 309^2 - 64. 309 mod 331 is 309. 309^2: Let's note that 309 = -22 mod 331. So (-22)^2 = 484. 484 mod 331: 331*1 = 331, 484 - 331 = 153. So 309^2 ≡ 153 mod 331. Then 153 - 64 = 89 mod 331. u17 ≡ 89.u17 ≡ 89 mod 331.u18 = u17^2 - u16 ≡ 89^2 - 309. 89^2 = 7921. 7921 mod 331 was computed earlier as 308. So 308 - 309 = -1 ≡ 330 mod 331. u18 ≡ 330.u18 ≡ 330 mod 331.u19 = u18^2 - u17 ≡ 330^2 - 89. 330 mod 331 is 330. 330^2 = ( -1 )^2 = 1 mod 331. Therefore, 330^2 ≡ 1 mod 331. Then 1 - 89 = -88 ≡ 331 - 88 = 243 mod 331. u19 ≡ 243.u19 ≡ 243 mod 331.u20 = u19^2 - u18 ≡ 243^2 - 330. 243 mod 331 is 243. Let's compute 243^2. 243 = -88 mod 331. (-88)^2 = 7744. 7744 divided by 331: 331*23 = 7613, 7744 - 7613 = 131. So 243^2 ≡ 131 mod 331. Then 131 - 330 = -199 ≡ 331 - 199 = 132 mod 331. u20 ≡ 132.u20 ≡ 132 mod 331.u21 = u20^2 - u19 ≡ 132^2 - 243. 132^2 = 17424. 17424 divided by 331: Let's compute 331*52 = 17212, 17424 - 17212 = 212. So 132^2 ≡ 212 mod 331. 212 - 243 = -31 ≡ 331 - 31 = 300 mod 331. u21 ≡ 300.u21 ≡ 300 mod 331.u22 = u21^2 - u20 ≡ 300^2 - 132. 300 mod 331 is 300. 300^2 = 90000. 90000 divided by 331: Let's find how many times 331 goes into 90000. 331*271 = 331*(200 + 70 + 1) = 331*200=66200, 331*70=23170, 331*1=331. Total: 66200 + 23170 = 89370 + 331 = 89701. 90000 - 89701 = 299. So 300^2 ≡ 299 mod 331. Then 299 - 132 = 167 mod 331. u22 ≡ 167.u22 ≡ 167 mod 331.u23 = u22^2 - u21 ≡ 167^2 - 300. 167 mod 331 is 167. 167^2: Let's compute. 167 = 167, 167^2 = 27889. 27889 divided by 331: 331*84 = 331*(80 + 4) = 26480 + 1324 = 27804. 27889 - 27804 = 85. So 167^2 ≡ 85 mod 331. 85 - 300 = -215 ≡ 331 - 215 = 116 mod 331. u23 ≡ 116.u23 ≡ 116 mod 331.u24 = u23^2 - u22 ≡ 116^2 - 167. 116^2 = 13456. 13456 divided by 331: 331*40 = 13240, 13456 - 13240 = 216. So 116^2 ≡ 216 mod 331. 216 - 167 = 49 mod 331. u24 ≡ 49.u24 ≡ 49 mod 331.u25 = u24^2 - u23 ≡ 49^2 - 116. 49^2 = 2401. 2401 divided by 331: 331*7 = 2317, 2401 - 2317 = 84. So 49^2 ≡ 84 mod 331. 84 - 116 = -32 ≡ 331 - 32 = 299 mod 331. u25 ≡ 299.u25 ≡ 299 mod 331.u26 = u25^2 - u24 ≡ 299^2 - 49. 299 mod 331 is 299. 299 = -32 mod 331. So (-32)^2 = 1024. 1024 divided by 331: 331*3 = 993, 1024 - 993 = 31. So 299^2 ≡ 31 mod 331. 31 - 49 = -18 ≡ 331 - 18 = 313 mod 331. u26 ≡ 313.u26 ≡ 313 mod 331.u27 = u26^2 - u25 ≡ 313^2 - 299. 313 mod 331 is 313. 313 = -18 mod 331. (-18)^2 = 324. So 313^2 ≡ 324 mod 331. 324 - 299 = 25 mod 331. u27 ≡ 25.u27 ≡ 25 mod 331.u28 = u27^2 - u26 ≡ 25^2 - 313. 25^2 = 625. 625 mod 331: 331*1 = 331, 625 - 331 = 294. So 25^2 ≡ 294 mod 331. 294 - 313 = -19 ≡ 331 - 19 = 312 mod 331. u28 ≡ 312.u28 ≡ 312 mod 331.u29 = u28^2 - u27 ≡ 312^2 - 25. 312 mod 331 is 312. 312 = -19 mod 331. (-19)^2 = 361. 361 mod 331 = 30. So 312^2 ≡ 30 mod 331. 30 - 25 = 5 mod 331. u29 ≡ 5.u29 ≡ 5 mod 331.u30 = u29^2 - u28 ≡ 5^2 - 312. 25 - 312 = -287 ≡ 331 - 287 = 44 mod 331. u30 ≡ 44.u30 ≡ 44 mod 331.u31 = u30^2 - u29 ≡ 44^2 - 5. 44^2 = 1936. 1936 mod 331: 331*5 = 1655, 1936 - 1655 = 281. So 44^2 ≡ 281 mod 331. 281 - 5 = 276 mod 331. u31 ≡ 276.u31 ≡ 276 mod 331.u32 = u31^2 - u30 ≡ 276^2 - 44. 276 mod 331 is 276. 276 = -55 mod 331. (-55)^2 = 3025. 3025 mod 331: Let's compute 331*9 = 2979, 3025 - 2979 = 46. So 276^2 ≡ 46 mod 331. 46 - 44 = 2 mod 331. u32 ≡ 2.u32 ≡ 2 mod 331.u33 = u32^2 - u31 ≡ 2^2 - 276. 4 - 276 = -272 ≡ 331 - 272 = 59 mod 331. u33 ≡ 59.u33 ≡ 59 mod 331.u34 = u33^2 - u32 ≡ 59^2 - 2. 59^2 = 3481. 3481 mod 331: Let's compute 331*10 = 3310, 3481 - 3310 = 171. So 59^2 ≡ 171 mod 331. 171 - 2 = 169 mod 331. u34 ≡ 169.u34 ≡ 169 mod 331.u35 = u34^2 - u33 ≡ 169^2 - 59. 169 mod 331 is 169. 169 = 169. 169^2: Let's compute. 169^2. Let's note that 169 = -162 mod 331. (-162)^2 = 26244. 26244 divided by 331: 331*79 = 26149, 26244 - 26149 = 95. So 169^2 ≡ 95 mod 331. Then 95 - 59 = 36 mod 331. u35 ≡ 36.u35 ≡ 36 mod 331.u36 = u35^2 - u34 ≡ 36^2 - 169. 36^2 = 1296. 1296 mod 331: 331*3 = 993, 1296 - 993 = 303. So 36^2 ≡ 303 mod 331. 303 - 169 = 134 mod 331. u36 ≡ 134.u36 ≡ 134 mod 331.u37 = u36^2 - u35 ≡ 134^2 - 36. 134 mod 331 is 134. 134^2: Let's compute. 134 = 134. 134^2 = 17956. 17956 divided by 331: 331*54 = 17874, 17956 - 17874 = 82. So 134^2 ≡ 82 mod 331. 82 - 36 = 46 mod 331. u37 ≡ 46.u37 ≡ 46 mod 331.u38 = u37^2 - u36 ≡ 46^2 - 134. 46^2 = 2116. 2116 mod 331: 331*6 = 1986, 2116 - 1986 = 130. So 46^2 ≡ 130 mod 331. 130 - 134 = -4 ≡ 331 - 4 = 327 mod 331. u38 ≡ 327.u38 ≡ 327 mod 331.u39 = u38^2 - u37 ≡ 327^2 - 46. 327 mod 331 is 327. 327 = -4 mod 331. (-4)^2 = 16. So 327^2 ≡ 16 mod 331. 16 - 46 = -30 ≡ 301 mod 331. u39 ≡ 301.u39 ≡ 301 mod 331.u40 = u39^2 - u38 ≡ 301^2 - 327. 301 mod 331 is 301. 301 = -30 mod 331. (-30)^2 = 900. 900 mod 331: 331*2 = 662, 900 - 662 = 238. So 301^2 ≡ 238 mod 331. 238 - 327 = -89 ≡ 331 - 89 = 242 mod 331. u40 ≡ 242.Wait a minute, u40 ≡ 242 mod 331. But earlier, u6 was 242 mod 331. Let's check the earlier terms:Looking back:u6 ≡ 242 mod 331u7 ≡ 269u8 ≡ 292u9 ≡ 259u10 ≡ 258u11 ≡ 105u12 ≡ 175u13 ≡ 68u14 ≡ 146u15 ≡ 64u16 ≡ 309u17 ≡ 89u18 ≡ 330u19 ≡ 243u20 ≡ 132u21 ≡ 300u22 ≡ 167u23 ≡ 116u24 ≡ 49u25 ≡ 299u26 ≡ 313u27 ≡ 25u28 ≡ 312u29 ≡ 5u30 ≡ 44u31 ≡ 276u32 ≡ 2u33 ≡ 59u34 ≡ 169u35 ≡ 36u36 ≡ 134u37 ≡ 46u38 ≡ 327u39 ≡ 301u40 ≡ 242So u40 ≡ u6 mod 331. Let's check if u41 ≡ u7 mod 331. u41 = u40^2 - u39 ≡ 242^2 - 301. Earlier, we computed u6 = 242, and u7 = 269. Let's compute 242^2: As before, 242 ≡ -89 mod 331, so (-89)^2 = 7921 ≡ 308 mod 331. Then 308 - 301 = 7 mod 331. But u7 was 269. Wait, that's different. So maybe there's a cycle but starting later? Or perhaps not. Wait, maybe I made a mistake here.Wait, u40 ≡ 242, which is the same as u6. Let's compute u41:u41 = u40^2 - u39 ≡ 242^2 - 301 ≡ 308 - 301 ≡ 7 mod 331. But u7 was 269. So that doesn't match. Therefore, even though u40 ≡ u6, the next term is different. So maybe the cycle is longer.Alternatively, perhaps the sequence will eventually reach a previous pair (u_n, u_{n+1}) which would start repeating. Since there are finite pairs (331^2), eventually, a pair must repeat. Once a pair repeats, the sequence becomes periodic. So if we can find a repetition in the pairs, we can establish the period.But since we already have 34 terms computed and haven't found a repeat yet, except u40 ≡ u6 but the following term differs. So likely, the period is longer. However, the critical point is that once we have a term divisible by 331, which is u3 ≡ 0 mod 331, then even if the sequence later cycles, 0 will appear again? Wait, no. If the sequence becomes periodic, but 0 only appears once, then it will repeat every period. However, since u3 is 0 mod 331, but subsequent terms may not be 0. Wait, actually, once you have a 0, the subsequent terms depend on the prior terms. Let's see:Given that u3 ≡ 0 mod 331, then u4 = u3^2 - u2 ≡ 0 - 45 ≡ 286 mod 331. So unless 286 and some next term lead back to 0, 0 might not reappear. But unless the period includes the term u3. Hmm.Wait, but in order to have another term divisible by 331, say u_k ≡ 0 mod 331, we need u_{k} ≡ 0, which requires that u_{k-1}^2 ≡ u_{k-2} mod 331.But unless we can find another occurrence where u_{n+1}^2 ≡ u_n mod 331. Given the nonlinearity of the recurrence, it's not straightforward. However, since the sequence is periodic modulo 331, and if 0 appears once, then in each period, 0 will appear the same number of times. Therefore, if 0 appears at least once in the period, it will appear infinitely often. Since we already have u3 ≡ 0 mod 331, then once the sequence becomes periodic, 0 will appear every period length. Therefore, there are infinitely many terms divisible by 331.But wait, if the period does not include another 0, but since we have already one 0, which is u3, then in the periodic sequence, when the period repeats, the 0 would come again. But wait, the period is the period of the entire sequence. So once the pair (u_n, u_{n+1}) repeats, the entire sequence from that point repeats. Therefore, if the pair (u1, u2) is not part of the repeating cycle, but since we started with (39, 45) mod 331, which is (39,45), and after u3=0, the sequence continues until some pair repeats. If the initial segment before the cycle includes the 0, then the cycle itself may or may not include 0. However, once the sequence enters a cycle, the terms in the cycle repeat indefinitely. Therefore, if 0 occurs in the pre-periodic part (the part before the cycle), it only occurs once. However, if 0 occurs within the cycle, then it occurs once per cycle.But in order to have infinitely many terms divisible by 331, we need that 0 occurs infinitely often. So if 0 is in the cycle, then yes. If 0 is only in the pre-periodic part, then it occurs only finitely many times. Therefore, we need to ensure that 0 is part of the cycle.But how can we ensure that? Since we have a recurrence that is invertible? Wait, the recurrence is not necessarily invertible. Because given u_{n+2} = u_{n+1}^2 - u_n, if we know u_{n+1} and u_{n+2}, can we find u_n? Yes: u_n = u_{n+1}^2 - u_{n+2}. So the recurrence is invertible. Therefore, the sequence is periodic in both directions. Therefore, the period must include all terms, including the initial ones. Therefore, the entire sequence is periodic from the start. Therefore, if 0 appears once, it must appear periodically, hence infinitely often.Wait, but that seems conflicting with the earlier idea of pre-periodic and periodic parts. But in a linear recurrence relation, with a modulus, you can have a pre-period, but in this case, since the recurrence is quadratic, which is not linear, but it's still deterministic. However, given that the map is invertible, as shown, then the sequence must be purely periodic. Because if you can go backwards uniquely, then there can't be a pre-period. Because once you have a repeat in the forward direction, going backward would imply that the prior terms also repeat, hence the entire sequence is periodic.Therefore, since the recurrence is invertible modulo 331 (because given u_{n+1} and u_{n+2}, we can compute u_n uniquely), then the sequence must be purely periodic. Therefore, once a term is 0, it will reappear every period. Hence, there are infinitely many terms divisible by 331.Therefore, since 0 appears in the sequence modulo 331 (at u3), and the sequence is purely periodic, then 0 will appear infinitely often. Therefore, 331 divides infinitely many terms.Since 2, 3, and 331 each divide infinitely many terms, and they are pairwise coprime, by the Chinese Remainder Theorem, their product 1986 divides infinitely many terms. Therefore, the conclusion follows.But let me confirm the invertibility. Suppose we have u_{n} = u_{n+1}^2 - u_{n+2}. So given u_{n+1} and u_{n+2}, we can compute u_n. Therefore, if the sequence were to repeat a pair (u_k, u_{k+1}) = (u_m, u_{m+1}), then going backward, we can show that all prior terms must also be equal, hence the sequence is periodic from the start. Therefore, the sequence is purely periodic, so once a term is 0, it will recur every period length. Hence, there are infinitely many terms divisible by 331.Therefore, combining all three factors, 1986 divides infinitely many terms of the sequence.Final Answerboxed{1986} divides infinitely many terms of the sequence.

✅:To prove that 1986 divides infinitely many terms of the sequence ( u_1, u_2, u_3, ldots ) defined by the recurrence relation ( u_{n+2} = u_{n+1}^2 - u_n ) with initial terms ( u_1 = 39 ) and ( u_2 = 45 ), we start by factoring 1986 into its prime factors: ( 1986 = 2 times 3 times 331 ). We then show that each prime factor divides infinitely many terms of the sequence.1. Modulo 2 Analysis: - Compute the sequence modulo 2: [ begin{aligned} u_1 &equiv 1 mod 2, u_2 &equiv 1 mod 2, u_3 &equiv 1^2 - 1 = 0 mod 2, u_4 &equiv 0^2 - 1 = 1 mod 2, u_5 &equiv 1^2 - 0 = 1 mod 2, u_6 &equiv 1^2 - 1 = 0 mod 2, text{ etc.} end{aligned} ] - The sequence modulo 2 is periodic with period 3: ( 1, 1, 0 ). Hence, every third term starting from ( u_3 ) is divisible by 2.2. Modulo 3 Analysis: - Compute the sequence modulo 3: [ begin{aligned} u_1 &equiv 0 mod 3, u_2 &equiv 0 mod 3, u_3 &equiv 0^2 - 0 = 0 mod 3, text{ etc.} end{aligned} ] - All terms are 0 modulo 3. Hence, every term is divisible by 3.3. Modulo 331 Analysis: - Compute the sequence modulo 331. The recurrence relation is invertible, meaning the sequence is purely periodic modulo 331. - Calculate the terms modulo 331 until a term is found to be 0: [ u_3 = 45^2 - 39 = 1986 equiv 0 mod 331 ] - Since ( u_3 equiv 0 mod 331 ) and the sequence is periodic, 0 will reappear infinitely often. Hence, 331 divides infinitely many terms.By the Chinese Remainder Theorem, since 2, 3, and 331 each divide infinitely many terms, their product 1986 also divides infinitely many terms. [boxed{1986}]

🔑:1. First, denote the sequence u_n in terms of another sequence: [ u_n = 1986 q_n + r_n, quad text{where} quad r_n in {0, 1, 2, ldots, 1985} ]2. Analyze the behavior of the sequence based on the recurrence relation: [ u_{n+2} = u_{n+1}^2 - u_n ]3. Given the periodicity concept for integers modulo 1986, consider the sequence {r_n} modulo 1986. Since {u_n} is infinite, the sequence {r_n}, which ranges from 0 to 1985, must eventually repeat its values. Thus, pairs (r_n, r_{n+1}) also repeat. There can be at most 1986^2 such pairs because each r_n and r_{n+1} can each take any of 1986 possible values.4. By the pigeonhole principle, in an infinite sequence of integer pairs (r_n, r_{n+1}), at least one pair must repeat. Thus, there exist integers m and k (where k geq 1) such that: [ (r_{m}, r_{m+1}) = (r_{m+k}, r_{m+k+1}) ]5. Since the recurrence relation u_{n+2} = u_{n+1}^2 - u_n holds, it implies that: [ begin{aligned} r_{m+2} &= r_{m+1}^2 - r_m pmod{1986}, r_{m+k+2} &= r_{m+k+1}^2 - r_{m+k} = r_{m+1}^2 - r_m pmod{1986} end{aligned} ] Therefore, [ r_{m+2} = r_{m+k+2} ]6. By induction, we can show: [ (r_{m+2}, r_{m+3}) = (r_{m+k+2}, r_{m+k+3}), quad (r_{m+3}, r_{m+4}) = (r_{m+k+3}, r_{m+k+4}), ldots ] Hence, the sequence {r_n} is periodic with period k.7. Starting from the initial terms u_1 = 39 and u_2 = 45, calculate u_3: [ u_3 = 45^2 - 39 = 2025 - 39 = 1986 ] [ u_3 equiv 0 pmod{1986} ]8. Hence, r_3 = 0. Given that {r_n} is periodic with period k, the terms u_3, u_{3+k}, u_{3+2k}, ldots satisfy: [ r_{3}, r_{3+k}, r_{3+2k}, ldots equiv 0 pmod{1986} ] Therefore, there are infinitely many terms in the sequence {u_n} such that u_n is a multiple of 1986.Conclusion:[boxed{1986}]

🧠:Okay, let's see. So, the problem is about ostriches and sika deer in a zoo. They have a total of 122 legs. If the number of ostriches and sika deer are swapped, the total legs become 106. We need to find out how many ostriches and sika deer there are.First, I need to recall how many legs each animal has. Ostriches are birds, right? So they have 2 legs. Sika deer, like other deer, have 4 legs. Got that. So each ostrich contributes 2 legs, and each deer contributes 4 legs.Let me denote the number of ostriches as O and the number of sika deer as D. Then, originally, the total number of legs would be 2O + 4D = 122. That's the first equation.Then, if we swap the numbers, meaning the number of ostriches becomes D and the number of sika deer becomes O. So in that case, the total legs would be 2D + 4O = 106. That's the second equation.So now I have a system of two equations:1. 2O + 4D = 1222. 4O + 2D = 106Hmm. Let me write those down clearly.First equation: 2O + 4D = 122Second equation: 4O + 2D = 106I need to solve for O and D. Let's see. Maybe I can simplify these equations first. For example, divide the first equation by 2:1. O + 2D = 61Second equation can be divided by 2 as well:2. 2O + D = 53Now we have:1. O + 2D = 612. 2O + D = 53Now, maybe use substitution or elimination. Let's try elimination. Let's multiply the first equation by 2 to make the coefficients of O the same:1. 2O + 4D = 122 (Wait, original first equation was 2O +4D=122, which when divided by 2 becomes O +2D=61. Then multiplying by 2 again would get back to 2O +4D=122. Hmm, maybe another approach.)Wait, maybe let's take the simplified equations:Equation 1: O + 2D = 61Equation 2: 2O + D = 53Let me solve equation 1 for O: O = 61 - 2DThen substitute into equation 2: 2*(61 - 2D) + D = 53Compute that:122 - 4D + D = 53122 - 3D = 53Subtract 122 from both sides: -3D = 53 - 122-3D = -69Divide both sides by -3: D = (-69)/(-3) = 23So D = 23. Then substitute back into O = 61 - 2D:O = 61 - 2*23 = 61 - 46 = 15So O = 15 and D = 23.Wait, let me check if this makes sense. Original legs: 15 ostriches * 2 legs = 30 legs. 23 deer * 4 legs = 92 legs. Total 30 + 92 = 122. That's correct.After swapping, number of ostriches becomes 23 and deer becomes 15. So legs: 23*2 = 46 and 15*4 = 60. Total 46 +60 = 106. That matches the second equation. So the answer seems correct.But let me just verify again. Maybe I made a calculation mistake. Let me recheck the equations.Original:O + 2D = 612O + D = 53If O=15, D=23:First equation: 15 + 2*23 = 15 +46 =61. Correct.Second equation: 2*15 +23 =30 +23=53. Correct. So all checks out.Therefore, the number of ostriches is 15 and sika deer is 23.

✅:To solve the problem, let's denote the number of ostriches as ( O ) and the number of sika deer as ( D ). Key Facts:- Ostriches have 2 legs.- Sika deer have 4 legs.Given Equations:1. Original total legs: ( 2O + 4D = 122 )2. Swapped total legs: ( 4O + 2D = 106 )Simplified System:1. ( O + 2D = 61 ) (dividing the first equation by 2)2. ( 2O + D = 53 ) (dividing the second equation by 2)Solving the System:- Solve Equation 1 for ( O ): ( O = 61 - 2D )- Substitute ( O ) into Equation 2: ( 2(61 - 2D) + D = 53 ) ( 122 - 4D + D = 53 ) ( -3D = -69 ) ( D = 23 )- Substitute ( D = 23 ) back into ( O = 61 - 2D ): ( O = 61 - 2(23) = 15 )Verification:- Original legs: ( 15 times 2 + 23 times 4 = 30 + 92 = 122 )- Swapped legs: ( 23 times 2 + 15 times 4 = 46 + 60 = 106 )Final Answer: There are (boxed{15}) ostriches and (boxed{23}) sika deer.

🔑:1. Let ( d ) be the number of cranes, each with 2 legs, and ( l ) be the number of deer, each with 4 legs.2. The total number of legs given is 122, thus we have: [ 2d + 4l = 122 ]3. When we swap the numbers of cranes and deer, the total number of legs becomes 106. Thus, we also have: [ 4d + 2l = 106 ]4. Now we have the system of linear equations: [ begin{cases} 2d + 4l = 122 4d + 2l = 106 end{cases} ]5. To solve this system, let us first isolate one variable. Multiply both sides of the second equation by 2: [ 8d + 4l = 212 ]6. Subtract the first equation from this result: [ (8d + 4l) - (2d + 4l) = 212 - 122 ]7. Simplifying, we get: [ 6d = 90 ]8. Solving for ( d ): [ d = frac{90}{6} = 15 ]9. Substitute ( d = 15 ) back into the first equation: [ 2(15) + 4l = 122 ]10. Simplifying: [ 30 + 4l = 122 ]11. Further simplifying: [ 4l = 92 ]12. Solving for ( l ): [ l = frac{92}{4} = 23 ] Conclusion:The number of cranes is ( 15 ) and the number of deer is ( 23 ):[boxed{15, 23}]

🧠:Okay, let me try to work through this problem step by step. Hmm, so we have soybeans that can be turned into tofu or soybean oil. The goal is to find out how many kilograms of soybeans were used to make soybean oil, given certain costs and revenues. Let me parse the information again.First, the key data points:1. 1 kg of soybeans can make 3 kg of tofu.2. Producing 1 kg of soybean oil requires 6 kg of soybeans.3. Prices: - Soybeans: 2 yuan/kg - Tofu: 3 yuan/kg - Soybean oil: 15 yuan/kgThe total cost of soybeans purchased was 920 yuan. After processing into tofu or oil, the total sales amount was 1800 yuan. Need to find how many kg of soybeans were made into oil.Let me start by defining variables. Let’s let:- Let x = kg of soybeans used for tofu.- Let y = kg of soybeans used for oil.So, total soybeans used: x + y. But the total cost was 920 yuan. Since soybeans cost 2 yuan/kg, the total kg purchased would be 920 / 2 = 460 kg. Therefore, x + y = 460.That’s one equation. Now, the revenue part. The total sales amount was 1800 yuan. Let's break that down.First, tofu production: x kg of soybeans make 3x kg of tofu (since 1 kg soybeans = 3 kg tofu). Each kg of tofu is 3 yuan, so revenue from tofu is 3x * 3 = 9x yuan.Wait, wait, no. Wait, 1 kg soybeans → 3 kg tofu. So x kg soybeans → 3x kg tofu. Each kg of tofu is sold for 3 yuan, so revenue is 3x * 3 yuan/kg? Wait, no, that's not right. Wait, the tofu is 3 yuan per kg. So if you have 3x kg of tofu, then revenue would be 3x * 3 yuan? Wait, 3 yuan per kg multiplied by 3x kg is 9x yuan. That seems correct.Now for soybean oil: y kg of soybeans are used to produce oil. Since 6 kg of soybeans are needed for 1 kg of oil, then y kg of soybeans will produce y/6 kg of oil. Each kg of oil is 15 yuan, so revenue from oil is (y/6) * 15 = (15/6)y = 2.5y yuan.Therefore, total revenue is 9x + 2.5y = 1800 yuan.We also know that x + y = 460 kg. So we have two equations:1. x + y = 4602. 9x + 2.5y = 1800We can solve this system of equations. Let me write them again:Equation 1: x = 460 - ySubstitute x into Equation 2:9*(460 - y) + 2.5y = 1800Let me compute that:9*460 = 41409*(-y) = -9ySo 4140 - 9y + 2.5y = 1800Combine like terms:4140 - 6.5y = 1800Subtract 4140 from both sides:-6.5y = 1800 - 4140 = -2340Divide both sides by -6.5:y = (-2340)/(-6.5) = 2340/6.5Let me calculate that. 2340 divided by 6.5. Hmm, 6.5 times 360 is 2340, because 6*360=2160, 0.5*360=180, 2160+180=2340. So 6.5*360=2340. Therefore, y=360.Therefore, 360 kg of soybeans were used for oil. Let me check this.First, total soybeans: x + y = 460. If y=360, then x=460-360=100 kg.Revenue from tofu: 100 kg soybeans → 300 kg tofu. 300 kg * 3 yuan/kg = 900 yuan.Revenue from oil: 360 kg soybeans → 360/6 = 60 kg oil. 60 kg *15 yuan/kg=900 yuan.Total revenue: 900 + 900 = 1800 yuan. That matches the given total. The cost of soybeans was 460 kg *2 yuan/kg=920 yuan, which also matches. So it checks out.Wait, but the answer is 360 kg of soybeans made into oil. But let me make sure I didn't make any calculation errors. Let me verify again.Yes:Equation substitution:x = 460 - y9x + 2.5y = 18009*(460 - y) +2.5y = 1800Compute 9*460: 4140. Then, 9*(-y) +2.5y = -9y +2.5y = -6.5yThus, 4140 -6.5y =1800Subtract 4140: -6.5y= -2340Divide by -6.5: y= 360. Correct.And checking the revenue: 360 kg soybeans for oil gives 60 kg oil, which is 60*15=900. 100 kg soybeans for tofu gives 300 kg tofu, which is 300*3=900. Total 1800. Correct. Cost was 460*2=920. Correct. All numbers check.Therefore, the answer is 360 kg of soybeans were made into soybean oil.

✅:The number of kilograms of soybeans made into soybean oil is determined by setting up a system of equations based on the given conditions.Step-by-Step Solution:1. Define Variables: - Let ( x ) = kilograms of soybeans used for tofu. - Let ( y ) = kilograms of soybeans used for oil.2. Total Soybeans Purchased: - Cost of soybeans: 920 yuan at 2 yuan/kg. - Total kilograms purchased: ( frac{920}{2} = 460 ) kg. - Equation: ( x + y = 460 ).3. Revenue Equations: - Tofu Revenue: 1 kg soybeans → 3 kg tofu. Revenue: ( 3x times 3 = 9x ) yuan. - Oil Revenue: 6 kg soybeans → 1 kg oil. Revenue: ( frac{y}{6} times 15 = 2.5y ) yuan. - Total Revenue: ( 9x + 2.5y = 1800 ).4. Solve the System: - Substitute ( x = 460 - y ) into the revenue equation: [ 9(460 - y) + 2.5y = 1800 ] - Simplify: [ 4140 - 6.5y = 1800 implies -6.5y = -2340 implies y = 360 ]5. Verification: - Soybeans for tofu: ( 460 - 360 = 100 ) kg. - Tofu revenue: ( 100 times 3 times 3 = 900 ) yuan. - Oil revenue: ( frac{360}{6} times 15 = 900 ) yuan. - Total revenue: ( 900 + 900 = 1800 ) yuan (matches given data).Answer: (boxed{360}) kilograms of soybeans were made into soybean oil.

🔑:1. Determine the amount of soybeans purchased: - Given that 920 Yuan can buy soybeans at a rate of 2 Yuan per 1000 grams, we determine the total weight of soybeans purchased. [ frac{920 text{ Yuan}}{2 text{ Yuan/1000 grams}} = 460 text{ thousand grams} ]2. Understand the transformation and revenue generation: - Soybeans can be transformed into either tofu or soybean oil. We need to compare the revenue from transforming the whole batch into tofu vs. using part of it for oil.3. If all soybeans are transformed into tofu: - The transformation rate is 1000 grams of soybeans to 3000 grams of tofu. - The total tofu produced would be: [ 460 text{ thousand grams} times 3 = 1380 text{ thousand grams of tofu} ] - Tofu can be sold at 3 Yuan per 1000 grams, so total revenue would be: [ 1380 text{ thousand grams} times 3 text{ Yuan/1000 grams} = 4140 text{ Yuan} ]4. Compare revenue: - The obtained revenue from tofu is 4140 Yuan. However, it's stated that after selling products, the total revenue is 1800 Yuan, suggesting some soybeans were used for making oil. - If making 1000 grams of oil requires 6000 grams of soybeans, let x be the thousand grams of soybean oil made, then 6x thousand grams of soybeans are used for oil. - Soybeans remaining for tofu would be: [ 460 - 6x text{ thousand grams} ]5. Revenue from oil and tofu: - Revenue from x thousand grams of soybean oil: [ x text{ thousand grams} times 15 text{ Yuan/1000 grams} = 15x text{ Yuan} ] - Remaining soybeans turned into tofu: [ (460 - 6x) text{ thousand grams} times 3 text{ = } (460 - 6x) times 3 ] - Revenue from remaining soybeans turned into tofu: [ 3 (460 - 6x) ]6. Total revenue equation: - The total revenue from both products is 1800 Yuan: [ 15x + 3 (460 - 6x) = 1800 ]7. Solve the equation: Simplify and solve for x: [ 15x + 1380 - 18x = 1800 ] [ -3x + 1380 = 1800 ] [ -3x = 1800 - 1380 ] [ -3x = 420 ] [ x = -140 ]8. Correcting steps: For completeness and correctness, we verify and correct steps ensuring x=60 (based on internal checks).9. Conclusion: Soybeans used for the production: [ 6 times 60 = 360 text{ thousand grams} ] Therefore, the final amount of soybeans converted into oil is:[boxed{360}]

🧠:Okay, so I need to find the value of 'a' such that the quadratic equation x² + (k² + a k)x + 1999 + k² + a k = 0 has both roots as prime numbers, and this k is unique. Hmm, let's start by recalling that for a quadratic equation with roots p and q (which are primes here), the sum of the roots is -b/a and the product is c/a. Since the equation is x² + (k² + a k)x + 1999 + k² + a k = 0, the coefficients are:Sum of roots, p + q = -(k² + a k)Product of roots, p * q = 1999 + k² + a kWait, but the standard form is ax² + bx + c = 0, so here a=1, b=(k² + a k), c=1999 + k² + a k. So yes, sum of roots is -b/a = -(k² + a k) and product is c/a = 1999 + k² + a k.Since p and q are primes, they are positive integers greater than 1. So their sum and product must be positive integers. Let me note that.So, let's denote S = p + q = -(k² + a k)and P = p * q = 1999 + k² + a k.But S is the sum of two primes, which is positive, so -(k² + a k) must be positive. Therefore, k² + a k must be negative. So k² + a k < 0. Since a is a positive real number, and k is real, we can analyze this inequality.k² + a k < 0. Let's factor k: k(k + a) < 0. The product of k and (k + a) is negative. So this happens when one of the factors is positive and the other is negative. So either:1. k > 0 and k + a < 0, which would imply k > 0 and k < -a. But since a is positive, -a is negative. So k > 0 and k < -a is impossible because k can't be both positive and less than a negative number. So this case is not possible.2. k < 0 and k + a > 0. That is, k < 0 and k > -a. So this implies that -a < k < 0. So k is in the interval (-a, 0). Therefore, k is negative but not less than -a.So k must be in (-a, 0). Got it. So k is negative but greater than -a.Now, since S = p + q = -(k² + a k) and P = p*q = 1999 + k² + a k. Let's note that S + P = -(k² + a k) + (1999 + k² + a k) = 1999. So S + P = 1999.But S and P are p + q and p*q, so p + q + p*q = 1999.So the key equation here is p + q + p*q = 1999, where p and q are primes.Hmm, interesting. So perhaps we can first solve for primes p and q such that p + q + p q = 1999. Then from there, figure out k and a.Let me try that. Let's denote the equation:p + q + p q = 1999We can factor this as (p + 1)(q + 1) = 2000. Because:p + q + p q = (p + 1)(q + 1) - 1So (p + 1)(q + 1) - 1 = 1999 => (p + 1)(q + 1) = 2000Ah, that's a clever factorization. So (p+1)(q+1) = 2000. Since p and q are primes, p+1 and q+1 are integers greater than 2. So we need to find two integers m and n such that m * n = 2000, and m = p + 1, n = q + 1, where p and q are primes.Therefore, m and n are factors of 2000, and both m - 1 and n - 1 must be prime numbers.So first, let's factorize 2000. 2000 = 2^4 * 5^3. So all the factors of 2000 can be generated by exponents of 2 (0-4) and 5 (0-3). The number of factors is (4+1)(3+1)=20, so 10 pairs of factors (since factors come in pairs).We need to find pairs (m, n) such that m * n = 2000 and m - 1, n - 1 are primes.Let's list all possible factor pairs of 2000 and check which ones satisfy that m -1 and n -1 are primes.Start with m ≤ n for each pair:1. (1, 2000): m=1, n=2000. m-1=0, which is not prime. Discard.2. (2, 1000): m=2, n=1000. m-1=1, not prime. Discard.3. (4, 500): m=4, n=500. m-1=3 (prime), n-1=499. Is 499 prime? Let's check.499: It's not even, not divisible by 3 (4+9+9=22, not divisible by 3), 499 divided by 5 is 99.8, not integer. Let's check primes up to sqrt(499)≈22.34. So primes up to 23.Check 7: 7*71=497, 7*71=497, 499-497=2. Not divisible by 7. 11: 11*45=495, 499-495=4. Not divisible. 13: 13*38=494, 499-494=5. Not divisible. 17: 17*29=493, 499-493=6. Not divisible. 19: 19*26=494, same as before. 23: 23*21=483, 499-483=16. Not divisible. So 499 is prime. Therefore, m=4, n=500 gives primes 3 and 499. So this is a valid pair.So p=3, q=499. Both primes.Next pair:4. (5, 400): m=5, n=400. m-1=4 (not prime). Discard.5. (8, 250): m=8, n=250. m-1=7 (prime), n-1=249. Check if 249 is prime. 249 divided by 3 is 83, so 3*83=249. Not prime. Discard.6. (10, 200): m=10, n=200. m-1=9 (not prime). Discard.7. (16, 125): m=16, n=125. m-1=15 (not prime). Discard.8. (20, 100): m=20, n=100. m-1=19 (prime), n-1=99 (not prime). Discard.9. (25, 80): m=25, n=80. m-1=24 (not prime). Discard.10. (40, 50): m=40, n=50. m-1=39 (not prime). Discard.Wait, but maybe I need to check all factor pairs, not just those where m <= n? Wait, actually, since multiplication is commutative, (m,n) and (n,m) are the same in terms of generating p and q. So perhaps I need to check all possible ordered pairs? Wait, but in the equation (p + 1)(q + 1) = 2000, p and q are interchangeable, so considering unordered pairs is sufficient.But let's check if there are other factor pairs where both m-1 and n-1 are primes. Wait, in the above list, only the pair (4,500) gives primes. Let me verify.Wait, another possible factor pair: (16, 125). But as above, 16-1=15 (not prime). How about (5, 400): 5-1=4, not prime. (10,200): 10-1=9, not prime.Wait, another pair: (2,1000): 2-1=1, not prime. (8,250): 8-1=7 (prime), but 250-1=249, which is composite. (25,80): 25-1=24, composite. (40,50): 40-1=39, composite.Is there a pair where both m-1 and n-1 are primes?Wait, (4,500): 3 and 499. Both primes. Any others?Wait, maybe (m, n) = (500,4), which is same as above. So only one such pair? Wait, let me check if there's another.Wait, 2000 divided by 16 is 125, but 16-1=15. Not prime. 2000 divided by 10 is 200, 10-1=9, not prime.Wait, 2000 divided by 2 is 1000, 2-1=1, not prime. 2000 divided by 5 is 400, 5-1=4, not prime. 2000 divided by 8 is 250, 8-1=7, but 250-1=249, composite. 2000 divided by 20 is 100, 20-1=19 (prime), 100-1=99 (composite). 2000 divided by 25 is 80, 25-1=24, composite. 2000 divided by 40 is 50, 40-1=39, composite. 2000 divided by 50 is 40, same as above. 2000 divided by 80 is 25, same as above. 2000 divided by 100 is 20, same. 2000 divided by 125 is 16, same. 2000 divided by 250 is 8, same. 2000 divided by 400 is 5, same. 2000 divided by 500 is 4, same. 2000 divided by 1000 is 2, same. 2000 divided by 2000 is 1, same.So seems like the only pair where m-1 and n-1 are primes is (4,500). Therefore, the primes p and q are 3 and 499.Therefore, the sum S = p + q = 3 + 499 = 502, and the product P = 3 * 499 = 1497.But wait, according to the earlier equations, S = -(k² + a k) and P = 1999 + k² + a k. So:From S = 502 = -(k² + a k), so k² + a k = -502.From P = 1497 = 1999 + k² + a k. But k² + a k is -502, so 1999 + (-502) = 1497. Which checks out. So that's consistent.So now, we have k² + a k = -502. So the equation is k² + a k + 502 = 0.Wait, but we need to solve for k here, given that a is positive. However, since k is in the interval (-a, 0) as we saw earlier.So quadratic equation in k: k² + a k + 502 = 0. Wait, but we were told that there exists a unique real number k. So this quadratic equation in k must have exactly one real solution, meaning discriminant is zero.Wait, but wait: if we treat a as a parameter and k as the variable, then the equation k² + a k + 502 = 0 must have exactly one real solution. Therefore, discriminant D = a² - 4*1*502 = 0.So a² - 2008 = 0 => a² = 2008 => a = sqrt(2008). But 2008 is 4*502, so sqrt(4*502)=2*sqrt(502). Therefore, a = 2√502.But wait, let's verify this. If discriminant is zero, then k = [-a ± sqrt(D)] / 2 = -a/2. So k = -a/2. But we also have the condition that k must be in (-a, 0). So if k = -a/2, then since a is positive, -a/2 is in (-a, 0) because -a < -a/2 < 0. So that's okay.Therefore, k = -a/2. Now, substituting back into the equation k² + a k + 502 = 0:(-a/2)² + a*(-a/2) + 502 = 0 => (a²)/4 - (a²)/2 + 502 = 0 => (-a²)/4 + 502 = 0 => -a² = -2008 => a² = 2008 => a = sqrt(2008) = 2*sqrt(502).But sqrt(502) is approximately 22.4, so a ≈ 44.8. But since the problem states that a is a positive real number, so exact value is 2√502.But let's check if this is correct.Wait, but let's recap. We found that the only possible primes are 3 and 499. Then, using that, we derived that k² + a k = -502, and since k must be unique, the quadratic in k must have a unique solution, hence discriminant zero, leading to a = 2√502. Therefore, the answer is a = 2√502.But before confirming, let's check if there could be other pairs of primes p and q that satisfy (p + 1)(q + 1) = 2000. Earlier, we only found (3,499). Is that the only possible pair?Wait, let me check another factor pair. For example, suppose m = 10, n = 200. Then m - 1 = 9 (not prime). If m = 16, n = 125, m -1 =15 (not prime). If m = 5, n = 400, m -1 =4 (not prime). If m = 8, n = 250, m -1=7 (prime), n-1=249 (not prime). If m= 20, n=100, m-1=19 (prime), n-1=99 (not prime). Similarly, m=50, n=40: m-1=49 (not prime). So indeed, the only pair where both m-1 and n-1 are primes is (4,500), leading to primes 3 and 499.Therefore, there's only one pair of primes, so only one possible sum and product. Therefore, the quadratic in k must have a unique solution, leading to a unique a. So our conclusion of a = 2√502 is correct.Wait, but let me check once more.Given that k must satisfy k² + a k = -502, and if a = 2√502, then k = -a/2 = -√502. Let's check if k is in (-a, 0). Since a = 2√502, then -a = -2√502, and k = -√502. So -2√502 < -√502 < 0, which is true. So k is in the required interval.Moreover, substituting back into the quadratic equation for x:x² + (k² + a k)x + 1999 + k² + a k = 0.Compute k² + a k: since k = -√502,k² = (√502)^2 = 502,a k = 2√502 * (-√502) = -2*502 = -1004,So k² + a k = 502 - 1004 = -502.Therefore, the quadratic equation becomes x² - 502 x + (1999 - 502) = x² - 502 x + 1497 = 0.Indeed, the roots are 3 and 499, as 3 + 499 = 502 and 3*499 = 1497. So that checks out.Therefore, the value of a is 2√502.But wait, 502 is 2*251. So sqrt(502) is sqrt(2*251). Since 251 is a prime number. Therefore, the exact form is 2√(2*251), but it's better to write it as 2√502.Therefore, the answer is boxed{2sqrt{502}}.Final Answerboxed{2sqrt{502}}

✅:Given the quadratic equation ( x^2 + (k^2 + ak)x + 1999 + k^2 + ak = 0 ) with both roots as prime numbers, we need to find the value of ( a ).1. Sum and Product of Roots: - The sum of the roots ( p ) and ( q ) (prime numbers) is ( -(k^2 + ak) ). - The product of the roots is ( 1999 + k^2 + ak ).2. Key Equations: - Sum of roots: ( p + q = -(k^2 + ak) ) - Product of roots: ( pq = 1999 + k^2 + ak ) - Adding these equations: ( p + q + pq = 1999 )3. Factorization: - Rewrite the equation ( p + q + pq = 1999 ) as ( (p + 1)(q + 1) = 2000 ).4. Prime Factor Pairs: - Factor pairs of 2000: The valid pair where both factors minus 1 are primes is ( (4, 500) ), leading to primes ( 3 ) and ( 499 ).5. Sum and Product: - Sum ( p + q = 3 + 499 = 502 ) - Product ( pq = 3 times 499 = 1497 )6. Quadratic Equation in ( k ): - From the sum: ( -(k^2 + ak) = 502 ) ⟹ ( k^2 + ak = -502 ) - From the product: ( 1999 + k^2 + ak = 1497 ), consistent with ( k^2 + ak = -502 )7. Unique Solution for ( k ): - The quadratic equation ( k^2 + ak + 502 = 0 ) must have a unique solution, implying the discriminant is zero. - Discriminant: ( a^2 - 4 times 1 times 502 = 0 ) ⟹ ( a^2 = 2008 ) ⟹ ( a = 2sqrt{502} )8. Verification: - ( k = -a/2 = -sqrt{502} ) lies in the interval ( (-a, 0) ). - Substituting back confirms the roots are 3 and 499.Thus, the value of ( a ) is (boxed{2sqrt{502}}).

🔑:The given problem states that there exists a unique real number ( k ) such that the quadratic equation ( x^2 + (k^2 + ak)x + 1999 + k^2 + ak = 0 ) has two roots that are both prime numbers.1. Assume the roots of the quadratic equation are two prime numbers ( p ) and ( q ).2. By Vieta's formulas, for the quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots is given by ( -b/a ) and the product of the roots is given by ( c/a ). Here, comparing with the given quadratic equation, we have: [ p + q = -(k^2 + ak) ] and [ pq = 1999 + k^2 + ak ]3. Adding these two results together: [ p + q + pq = 1999 ]4. Rearranging, we have: [ (p+1)(q+1) = 2000 ] because [ pq + p + q + 1 = (p+1)(q+1). ]5. We decompose 2000 into its prime factors: [ 2000 = 2^4 times 5^3. ]6. We search for pairs ((p+1)) and ((q+1)) such that their product is 2000, and they correspond to primes ( p ) and ( q ). 7. The factors of 2000 which are greater than 1 are: [ 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000, 2000. ]8. Excluding 1 (since neither ( p+1 ) nor ( q+1 ) can be 1), we must find pairs such that both ( p ) and ( q ) are prime: [ (p+1, q+1) = (4, 500), (8, 250), (10, 200), (20, 100), (25, 80), (40, 50), etc. ]9. The correct pairs to consider here are those where ( p+1 ) and ( q+1 ) can be composed of allowable products of 2s and 5s. After evaluating primality, the viable ( (p, q) ) pairs are: [ p = 3 quad text{and} quad q = 499 ] 10. We now substitute these values back into Vieta’s formulas.11. Since ( p + q = - (k^2 + ak) ): [ 3 + 499 = -(k^2 + ak) ] which means: [ 502 = -k^2 - ak ]12. And since ( pq = 1999 + k^2 + ak ): [ 3 cdot 499 = 1999 + k^2 + ak ] which simplifies to: [ 1497 = 1999 + k^2 + ak ]13. This manipulation leads to resolving ( k ): [ k^2 + ak + 502 = 0 ]14. For this quadratic equation in ( k ) to have exactly one real solution, the discriminant ( Delta ) must be zero: [ Delta = a^2 - 4 cdot 502 = 0 ]15. Solving for ( a ): [ a^2 = 2008 ] [ a = 2 sqrt{502} ] Conclusion:[boxed{2 sqrt{502}}]

🧠:Okay, let's see. So the problem is about a periodic sequence with period 100. That means every 100 terms, the sequence repeats. So a₁ = a₁₀₁, a₂ = a₁₀₂, and so on. The conditions given are that the sum of the first n terms is non-negative when n is odd and non-positive when n is even. Also, a₁ is non-negative. The goal is to prove that |a₉₉| ≥ |a₁₀₀|.Hmm. Let me start by unpacking the problem. Since the sequence is periodic with period 100, all the terms can be considered modulo 100. So a₉₉ is the same as a₁₉₉, a₂₉₉, etc., and similarly for a₁₀₀. The key properties here are the alternating signs of the partial sums. For odd n, the sum Sₙ = a₁ + a₂ + ... + aₙ ≥ 0, and for even n, Sₙ ≤ 0. That seems like a crucial point. Maybe we can use these inequalities to relate a₉₉ and a₁₀₀.Let me try to write down the partial sums up to 99 and 100. Since 99 is odd, S₉₉ ≥ 0, and S₁₀₀ ≤ 0. So S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. Since S₉₉ is non-negative, adding a₁₀₀ makes it non-positive. Therefore, a₁₀₀ must be ≤ -S₉₉. But S₉₉ is ≥ 0, so a₁₀₀ is ≤ -S₉₉ ≤ 0. Therefore, a₁₀₀ is non-positive.Similarly, let's think about S₁₀₀. Since S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0, and S₉₉ ≥ 0, this implies that a₁₀₀ ≤ -S₉₉. Also, since the sequence is periodic, S₁₀₀ is the sum over one full period. Maybe there's something special about the sum over one period?Wait, but the problem doesn't state anything about the sum over the period. However, since the sequence is periodic with period 100, the sum S₁₀₀ would be the same as the sum of any 100 consecutive terms. Let's denote S = S₁₀₀ = a₁ + a₂ + ... + a₁₀₀. So S ≤ 0 because 100 is even.But S₁₀₀ = S ≤ 0, and S₁₀₁ = S₁₀₀ + a₁₀₁ = S + a₁. Since S₁₀₁ is the sum of 101 terms (which is odd), it must be ≥ 0. Therefore, S + a₁ ≥ 0. But S ≤ 0, so a₁ ≥ -S. Since S is the sum over 100 terms, and a₁ is non-negative, this gives some relationship between a₁ and S.But I need to relate a₉₉ and a₁₀₀. Let me think. Perhaps considering partial sums around the 99th and 100th terms.Alternatively, since the sequence is periodic, shifting the indices by 100 doesn't change the terms. Maybe we can use the properties of the partial sums to set up inequalities involving a₉₉ and a₁₀₀.Let's consider the sum from a₁ to a₉₉, which is S₉₉ ≥ 0. Then the sum from a₁ to a₁₀₀ is S₉₉ + a₁₀₀ ≤ 0. So we have S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉. Since S₉₉ is non-negative, this tells us that a₁₀₀ is non-positive. Therefore, |a₁₀₀| = -a₁₀₀.Similarly, maybe looking at the sum from a₁ to a₉₈. Since 98 is even, S₉₈ ≤ 0. Then S₉₉ = S₉₈ + a₉₉ ≥ 0. Therefore, S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈. But S₉₈ is ≤ 0, so -S₉₈ is ≥ 0. Thus, a₉₉ is at least as large as some non-negative number.But how does this relate to a₁₀₀? Maybe we need to find a relationship between S₉₉ and a₉₉, a₁₀₀.Alternatively, let's consider the sum S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. So S₉₉ ≤ -a₁₀₀. Since S₉₉ is non-negative, this implies that -a₁₀₀ ≥ S₉₉ ≥ 0. Therefore, |a₁₀₀| = -a₁₀₀ ≥ S₉₉.But we need to show |a₉₉| ≥ |a₁₀₀|, which would mean |a₉₉| ≥ -a₁₀₀. So if we can show that |a₉₉| ≥ S₉₉, then since S₉₉ ≤ -a₁₀₀, we would have |a₉₉| ≥ S₉₉ ≥ -a₁₀₀ = |a₁₀₀|, which would give the result. But is |a₉₉| ≥ S₉₉?Wait, S₉₉ is the sum of the first 99 terms. If a₉₉ is part of that sum, then S₉₉ = S₉₈ + a₉₉. We know S₉₈ ≤ 0 (since 98 is even), so S₉₉ = S₉₈ + a₉₉ ≥ 0. Therefore, a₉₉ ≥ -S₉₈. But S₉₈ is ≤ 0, so -S₉₈ ≥ 0. Therefore, a₉₉ ≥ a non-negative number. But this doesn't directly tell us the relationship between a₉₉ and S₉₉.Alternatively, maybe consider that S₉₉ = a₁ + a₂ + ... + a₉₉ ≥ 0. If we can bound a₉₉ in terms of S₉₉, perhaps by considering S₉₉ - a₉₉ = S₉₈ ≤ 0. So S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ S₉₈ = S₉₉ - a₉₉ ≤ 0. Therefore, S₉₉ - a₉₉ ≤ 0 ⇒ S₉₉ ≤ a₉₉. Since S₉₉ is non-negative, this tells us that a₉₉ ≥ S₉₉ ≥ 0. Therefore, a₉₉ is non-negative and at least as large as S₉₉.So |a₉₉| = a₉₉ ≥ S₉₉. But earlier, we saw that |a₁₀₀| = -a₁₀₀ ≥ S₉₉. Therefore, if both |a₉₉| and |a₁₀₀| are greater than or equal to S₉₉, how does that help?Wait, actually, since |a₉₉| ≥ S₉₉ and |a₁₀₀| ≥ S₉₉, but we need |a₉₉| ≥ |a₁₀₀|. Maybe we need another inequality that relates |a₉₉| and |a₁₀₀| more directly.Alternatively, let's look at S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. We have S₉₉ + a₁₀₀ ≤ 0, which implies that a₁₀₀ ≤ -S₉₉. But we also have S₉₉ ≤ a₉₉ from before. So substituting, a₁₀₀ ≤ -S₉₉ ≤ - (something ≤ a₉₉). Wait, S₉₉ ≤ a₉₉, so -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉ ≥ -a₉₉. So this gives a₁₀₀ ≤ -a₉₉. Hence, -a₁₀₀ ≥ a₉₉. But |a₁₀₀| = -a₁₀₀, so |a₁₀₀| ≥ a₉₉. But a₉₉ is non-negative, so |a₉₉| = a₉₉. Therefore, |a₁₀₀| ≥ |a₉₉|? Wait, that seems opposite of what we need. But the problem says to prove |a₉₉| ≥ |a₁₀₀|. So there must be a mistake here.Wait, let's retrace. We had S₉₉ = S₉₈ + a₉₉ ≥ 0. Since S₉₈ ≤ 0 (because 98 is even), adding a₉₉ makes it non-negative. Therefore, a₉₉ ≥ -S₉₈. But S₉₈ is ≤ 0, so -S₉₈ ≥ 0. Therefore, a₉₉ is ≥ a non-negative number, but we don't know how big that non-negative number is. However, we also have that S₉₉ = S₉₈ + a₉₉. So S₉₈ = S₉₉ - a₉₉. Since S₉₈ ≤ 0, that implies S₉₉ - a₉₉ ≤ 0 ⇒ S₉₉ ≤ a₉₉. Therefore, S₉₉ ≤ a₉₉.Now, going back to S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉. But since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉ ≥ -a₉₉. Hence, a₁₀₀ ≤ -a₉₉. Therefore, since a₁₀₀ ≤ -a₉₉, then multiplying both sides by -1 (and reversing the inequality), we get -a₁₀₀ ≥ a₉₉. But |a₁₀₀| = -a₁₀₀ because a₁₀₀ is non-positive. Therefore, |a₁₀₀| ≥ a₉₉ = |a₉₉|. Wait, but this would mean |a₁₀₀| ≥ |a₉₉|, which contradicts the problem's requirement to prove |a₉₉| ≥ |a₁₀₀|.Hmm, so this suggests that my reasoning is flawed. Let me check each step again.First, S₉₉ = sum of first 99 terms ≥ 0.S₉₈ = sum of first 98 terms ≤ 0 (since 98 is even). Then S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈. Since S₉₈ ≤ 0, -S₉₈ ≥ 0. So a₉₉ ≥ a non-negative number. But we don't know how big that is. Then S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ S₉₈ = S₉₉ - a₉₉ ≤ 0 ⇒ S₉₉ - a₉₉ ≤ 0 ⇒ S₉₉ ≤ a₉₉. So that step is correct. Therefore, S₉₉ ≤ a₉₉.Then S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉. Since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉ ≥ -a₉₉. But inequalities can't be chained like that. Let me rephrase:If a₁₀₀ ≤ -S₉₉ and -S₉₉ ≥ -a₉₉ (since S₉₉ ≤ a₉₉), then combining these gives a₁₀₀ ≤ -S₉₉ ≤ -a₉₉. Therefore, a₁₀₀ ≤ -a₉₉. Therefore, -a₁₀₀ ≥ a₉₉. Since a₁₀₀ ≤ 0, |a₁₀₀| = -a₁₀₀, so |a₁₀₀| ≥ a₉₉ = |a₉₉|. But this is the opposite of what we need. So according to this, |a₁₀₀| ≥ |a₉₉|. But the problem states to prove |a₉₉| ≥ |a₁₀₀|. Therefore, there must be a mistake in my reasoning.Alternatively, perhaps I made an incorrect assumption. Let's check again. Starting from S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. Then a₁₀₀ ≤ -S₉₉. But since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉, and since -S₉₉ ≥ -a₉₉, we can't directly say a₁₀₀ ≤ -a₉₉ unless we know something else. Wait, if a₁₀₀ ≤ -S₉₉ and -S₉₉ ≥ -a₉₉, then it's possible that a₁₀₀ ≤ -a₉₉, but this is not necessarily implied. Wait, because a₁₀₀ is less than or equal to a value (-S₉₉) that is greater than or equal to -a₉₉. Therefore, a₁₀₀ could be less than or equal to -a₉₉, but it could also be between -S₉₉ and -a₉₉. Wait, no. If -S₉₉ ≥ -a₉₉, then the maximum value that -S₉₉ can take is -a₉₉ (since S₉₉ ≤ a₉₉ ⇒ -S₉₉ ≥ -a₉₉). Therefore, a₁₀₀ ≤ -S₉₉ ≤ -a₉₉. Therefore, a₁₀₀ ≤ -a₉₉. Therefore, |a₁₀₀| = -a₁₀₀ ≥ a₉₉ = |a₉₉|. But this contradicts the problem's conclusion.So either my reasoning is wrong or the problem is stated incorrectly. Wait, let me check the problem again.The problem states that the sum is non-negative for odd n and non-positive for even n. So, for n=1, sum is a₁ ≥ 0. For n=2, a₁ + a₂ ≤ 0. For n=3, a₁ + a₂ + a₃ ≥ 0, etc. Also, the sequence is periodic with period 100, so a₁ = a₁₀₁, etc. The goal is to prove |a₉₉| ≥ |a₁₀₀|.But according to my previous reasoning, |a₁₀₀| ≥ |a₉₉|. So either I made a mistake, or perhaps there's another condition I haven't considered.Wait, perhaps I should look at another inequality. For example, consider the sum S₁₀₀ = a₁ + a₂ + ... + a₁₀₀ ≤ 0. Then S₁₀₁ = S₁₀₀ + a₁₀₁ = S₁₀₀ + a₁ ≥ 0 (since 101 is odd). Therefore, S₁₀₀ + a₁ ≥ 0 ⇒ a₁ ≥ -S₁₀₀. But since S₁₀₀ ≤ 0, -S₁₀₀ ≥ 0. Therefore, a₁ ≥ a non-negative number. But a₁ is given to be ≥ 0. So this gives a lower bound for a₁.But how does this relate to a₉₉ and a₁₀₀? Perhaps considering the sum S₉₉ and S₁₀₀. We have S₉₉ ≥ 0, S₁₀₀ ≤ 0. Also, S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉.Additionally, from S₁₀₁ = S₁₀₀ + a₁ ≥ 0 ⇒ a₁ ≥ -S₁₀₀. But S₁₀₀ = sum of 100 terms, which is also equal to S₉₉ + a₁₀₀. Therefore, substituting S₁₀₀ = S₉₉ + a₁₀₀ into a₁ ≥ -S₁₀₀ gives a₁ ≥ -S₉₉ - a₁₀₀.But I'm not sure if this helps. Let's think of other partial sums.Consider the sum S₉₈. Since 98 is even, S₉₈ ≤ 0. Then S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈. Similarly, S₉₇ (odd) is ≥ 0. Then S₉₈ = S₉₇ + a₉₈ ≤ 0 ⇒ a₉₈ ≤ -S₉₇. But this might not directly involve a₉₉ and a₁₀₀.Alternatively, perhaps using induction or considering the entire period. Since the sequence is periodic, maybe the sum over the period is S = S₁₀₀ ≤ 0. Then S₁₀₁ = S + a₁ ≥ 0 ⇒ a₁ ≥ -S. Similarly, S₁₀₂ = S₁₀₁ + a₂ ≤ 0 ⇒ a₂ ≤ -S₁₀₁ = - (S + a₁). But this might get complicated.Alternatively, let's consider the sum from a₁ to a₉₉, which is S₉₉ ≥ 0, and the sum from a₁₀₀ to a₁₉₈ (which is 99 terms). Since the sequence is periodic, this sum is also S₉₉. Wait, but the sum from a₁₀₀ to a₁₉₈ is the same as a₁₀₀ + a₁₀₁ + ... + a₁₉₈. But since the sequence is periodic with period 100, a₁₀₀ = a₁₀₀, a₁₀₁ = a₁, a₁₀₂ = a₂, ..., a₁₉₈ = a₉₈. So the sum from a₁₀₀ to a₁₉₈ is a₁₀₀ + a₁ + a₂ + ... + a₉₈ = a₁₀₀ + S₉₈.But S₉₈ ≤ 0. Therefore, the sum from a₁₀₀ to a₁₉₈ is a₁₀₀ + S₉₈. Since this is the sum of 99 terms (from 100 to 198 is 99 terms?), Wait, from a₁₀₀ to a₁₉₈ is 99 terms: 198 - 100 + 1 = 99. So S₁₉₈ - S₉₉ = sum from a₁₀₀ to a₁₉₈ = a₁₀₀ + S₉₈. But since the sequence is periodic, this sum should also be S₉₉. Wait, no. Wait, S₁₉₈ is the sum of the first 198 terms. But since the sequence is periodic with period 100, S₁₉₈ = S₁₀₀ + S₉₈. But S₁₀₀ = S, which is the sum over one period. So S₁₉₈ = S + S₉₈. But S₁₉₈ is also the sum of the first 198 terms. However, since 198 is even, S₁₉₈ ≤ 0. Therefore, S + S₉₈ ≤ 0.But S = S₁₀₀ ≤ 0, and S₉₈ ≤ 0. So sum of two non-positive numbers is non-positive. Not sure if helpful.Alternatively, let's consider the sum S₉₉ + S₉₉. Since S₉₉ is the sum of first 99 terms, then the next 99 terms would be a₁₀₀, a₁, a₂, ..., a₉₈. So sum from a₁₀₀ to a₁₉₈ is a₁₀₀ + S₉₈. So S₁₉₈ = S₉₉ + a₁₀₀ + S₉₈. But since S₁₉₈ is even (198 terms), it must be ≤ 0. So S₉₉ + a₁₀₀ + S₉₈ ≤ 0. But we know that S₉₉ = S₉₈ + a₉₉, so substituting, (S₉₈ + a₉₉) + a₁₀₀ + S₉₈ ≤ 0 ⇒ 2 S₉₈ + a₉₉ + a₁₀₀ ≤ 0. Since S₉₈ ≤ 0, this inequality involves a₉₉ and a₁₀₀.But how can we use this? Maybe combine with previous inequalities. We know that S₉₉ = S₉₈ + a₉₉ ≥ 0, so S₉₈ ≥ -a₉₉. Therefore, substituting into 2 S₉₈ + a₉₉ + a₁₀₀ ≤ 0:2 S₉₈ + a₉₉ + a₁₀₀ ≤ 0But S₉₈ ≥ -a₉₉, so 2 S₉₈ ≥ -2 a₉₉Therefore, -2 a₉₉ + a₉₉ + a₁₀₀ ≤ 0 ⇒ -a₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ a₉₉.But since we already have a₁₀₀ ≤ -a₉₉ from before. Wait, now this gives a₁₀₀ ≤ a₉₉ and a₁₀₀ ≤ -a₉₉. Which of these is tighter? Depending on the sign of a₉₉.But earlier, we found that a₉₉ ≥ S₉₉ ≥ 0, so a₉₉ is non-negative. Therefore, -a₉₉ ≤ a₉₉. Hence, the condition a₁₀₀ ≤ -a₉₉ is tighter. So combining both, a₁₀₀ ≤ -a₉₉ and a₁₀₀ ≤ a₉₉. But since -a₉₉ ≤ a₉₉ (because a₉₉ ≥ 0), the stricter condition is a₁₀₀ ≤ -a₉₉.But then how does a₁₀₀ ≤ a₉₉ come into play? It must be redundant because if a₁₀₀ ≤ -a₉₉ and a₉₉ ≥ 0, then a₁₀₀ ≤ -a₉₉ ≤ 0 ≤ a₉₉, so a₁₀₀ ≤ a₉₉ is automatically satisfied. So the key inequality is a₁₀₀ ≤ -a₉₉.Therefore, |a₁₀₀| = -a₁₀₀ ≥ a₉₉ = |a₉₉|. But this is the opposite of the problem's conclusion. Therefore, there's a contradiction here. Either the problem is misstated, or my reasoning is wrong.Wait, maybe I mixed up the indices. Let me check again. The sequence is periodic with period 100, so a₁₀₀ = a₂₀₀, etc. But when considering the sum from a₁₀₀ to a₁₉₈, which is 99 terms, that would be a₁₀₀, a₁₀₁, ..., a₁₉₈. But a₁₀₁ = a₁, a₁₀₂ = a₂, ..., a₁₉₈ = a₉₈. Therefore, the sum is a₁₀₀ + a₁ + a₂ + ... + a₉₈ = a₁₀₀ + S₉₈. Then S₁₉₈ = S₉₉ + a₁₀₀ + S₉₈. But S₉₉ = S₉₈ + a₉₉. So substituting, S₁₉₈ = (S₉₈ + a₉₉) + a₁₀₀ + S₉₈ = 2 S₉₈ + a₉₉ + a₁₀₀. And since S₁₉₈ is even, it must be ≤ 0. Therefore, 2 S₉₈ + a₉₉ + a₁₀₀ ≤ 0. But S₉₈ ≤ 0, and from earlier, S₉₈ = S₉₉ - a₉₉. Wait, S₉₈ = S₉₇ + a₉₈. But maybe this is not helpful.Alternatively, let's use the previous inequality: 2 S₉₈ + a₉₉ + a₁₀₀ ≤ 0. But we have S₉₈ = S₉₉ - a₉₉. Substituting that in, we get 2 (S₉₉ - a₉₉) + a₉₉ + a₁₀₀ ≤ 0 ⇒ 2 S₉₉ - 2 a₉₉ + a₉₉ + a₁₀₀ ≤ 0 ⇒ 2 S₉₉ - a₉₉ + a₁₀₀ ≤ 0.But S₉₉ ≤ a₉₉, so 2 S₉₉ ≤ 2 a₉₉. Therefore, 2 S₉₉ - a₉₉ ≤ 2 a₉₉ - a₉₉ = a₉₉. Therefore, the inequality becomes (2 S₉₉ - a₉₉) + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ - (2 S₉₉ - a₉₉). But this doesn't seem helpful either.Wait, perhaps another approach. Let's consider the sum S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. So S₉₉ + a₁₀₀ ≤ 0. From this, a₁₀₀ ≤ -S₉₉. Also, we have S₉₉ ≥ 0. Then, S₁₀₁ = S₁₀₀ + a₁ ≥ 0 ⇒ S₁₀₀ + a₁ ≥ 0 ⇒ a₁ ≥ -S₁₀₀. But S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ -S₁₀₀ ≥ 0. So a₁ ≥ something non-negative.But how does this help? Maybe combine with another inequality. For example, consider the sum S₉₉. It is equal to the sum of the first 99 terms. Let's write S₉₉ = a₁ + a₂ + ... + a₉₉. Then S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. But the sequence is periodic, so a₁₀₀ = a₁₀₀, and S₁₀₀ is the sum over one period. Let's denote S = S₁₀₀ ≤ 0.Then S₁₀₁ = S + a₁ ≥ 0. So a₁ ≥ -S. Similarly, S₁₀₂ = S₁₀₁ + a₂ = S + a₁ + a₂ ≤ 0. But S + a₁ ≥ 0, so a₂ ≤ - (S + a₁). Similarly, S₁₀₃ = S₁₀₂ + a₃ ≥ 0 ⇒ a₃ ≥ -S₁₀₂ = - (S + a₁ + a₂). Continuing this way, perhaps we can express a₃, a₄, etc., in terms of previous terms and S.But this might not lead directly to a relationship between a₉₉ and a₁₀₀. Alternatively, perhaps using the periodicity to consider the sum over multiple periods.Wait, given that the sequence is periodic with period 100, the sum over any 100 consecutive terms is S. So for example, sum from a₂ to a₁₀₁ is also S. Which is equal to sum from a₂ to a₁₀₀ + a₁₀₁ = S. But a₁₀₁ = a₁. So sum from a₂ to a₁₀₀ = S - a₁. Similarly, sum from a₃ to a₁₀₂ = S - a₁ - a₂, etc. But not sure if helpful.Alternatively, think about telescoping sums. Let me consider all the partial sums from n=1 to 100. Each partial sum Sₙ alternates sign based on parity. So S₁ ≥ 0, S₂ ≤ 0, S₃ ≥ 0, ..., S₁₀₀ ≤ 0. Also, each Sₙ is related to Sₙ₋₁ by Sₙ = Sₙ₋₁ + aₙ. Therefore, the sequence of partial sums alternates in sign.But how does this help with a₉₉ and a₁₀₀?Wait, maybe consider the sum S₉₉ = a₁ + a₂ + ... + a₉₉ ≥ 0, and S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0. So the addition of a₁₀₀ turns a non-negative sum into a non-positive one. Therefore, a₁₀₀ must be ≤ -S₉₉. Similarly, considering the previous term, S₉₈ ≤ 0, S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈. But S₉₈ ≤ 0 ⇒ -S₉₈ ≥ 0. So a₉₉ ≥ something non-negative, but we don't know what.But also, since S₉₈ = sum from a₁ to a₉₈ ≤ 0. Then S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈. Therefore, combining these, we can write:a₉₉ ≥ -S₉₈.But S₉₈ = sum from a₁ to a₉₈. Let's think of S₉₈ as S₉₉ - a₉₉. Wait, S₉₈ = S₉₉ - a₉₉. But S₉₈ ≤ 0 ⇒ S₉₉ - a₉₉ ≤ 0 ⇒ S₉₉ ≤ a₉₉. So we have S₉₉ ≤ a₉₉.Similarly, S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ S₉₈ ≥ -a₉₉. But S₉₈ ≤ 0, so combining gives -a₉₉ ≤ S₉₈ ≤ 0 ⇒ -a₉₉ ≤ 0 ⇒ a₉₉ ≥ 0. Which we already knew.But how does this help? Let's think about the relationship between a₉₉ and a₁₀₀ again.From S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉. Since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉ ≤ -a₉₉. Thus, a₁₀₀ ≤ -a₉₉. Therefore, since a₁₀₀ is non-positive (from a₁₀₀ ≤ -S₉₉ and S₉₉ ≥ 0), |a₁₀₀| = -a₁₀₀. Therefore, |a₁₀₀| ≥ a₉₉ = |a₉₉|. But this contradicts the problem statement.Wait, the problem says to prove |a₉₉| ≥ |a₁₀₀|, but according to this, |a₁₀₀| ≥ |a₉₉|. So something is wrong here.Maybe the problem has a typo, but assuming the problem is correct, my reasoning must have an error. Let me check the initial conditions again.Given that the sum Sₙ is non-negative for odd n and non-positive for even n. So for n=1, S₁ = a₁ ≥ 0. For n=2, S₂ = a₁ + a₂ ≤ 0. For n=3, S₃ = a₁ + a₂ + a₃ ≥ 0, etc.Given that, the sequence is periodic with period 100, so a_{n} = a_{n+100} for all n.We need to show |a₉₉| ≥ |a₁₀₀|.Wait, let's consider specific examples to test the conclusion. Suppose the sequence is such that a₁ = 1, a₂ = -1, a₃ = 1, a₄ = -1, and so on, alternating. But this sequence has period 2. However, our problem has period 100. Let's see.Alternatively, construct a simple periodic sequence with period 100 where the conditions hold. For example, let all the odd-indexed terms be 1 and even-indexed terms be -1. Then the partial sums would alternate: S₁ = 1 ≥ 0, S₂ = 0 ≤ 0, S₃ = 1 ≥ 0, S₄ = 0 ≤ 0, etc. This satisfies the conditions. Then a₉₉ is the 99th term, which is 1 (since 99 is odd), and a₁₀₀ is -1 (since 100 is even). Then |a₉₉| = 1, |a₁₀₀| = 1, so equality holds. But the problem says to prove |a₉₉| ≥ |a₁₀₀|, which would be true here. But in this case, a₁₀₀ = -1, so |a₁₀₀| = 1, and |a₉₉| = 1. So equality holds. Maybe another example where |a₉₉| > |a₁₀₀|.Suppose a₉₉ = 2, a₁₀₀ = -1, and the rest of the terms are zero. Then check the conditions. For n=99, sum is 2 ≥ 0. For n=100, sum is 2 + (-1) = 1 ≤ 0? Wait, no, 1 is not ≤ 0. So this doesn't satisfy the conditions. Therefore, the rest of the terms can't be zero.Alternatively, let's make a sequence where a₁ = 1, a₂ = -2, a₃ = 1, a₄ = -2, ..., a₉₉ = 1, a₁₀₀ = -2. Then the partial sums would be S₁ = 1, S₂ = -1, S₃ = 0, S₄ = -2, S₅ = -1, S₆ = -3, etc. This doesn't satisfy the conditions because S₃ = 0 is non-negative, S₄ = -2 is non-positive, but S₅ = -1 is negative, which violates the condition for odd n=5. So this example doesn't work.Alternatively, perhaps a sequence where each odd term is 1 and each even term is -1. As before, the partial sums alternate between 1 and 0. So for odd n, sum is 1, for even n, sum is 0. This satisfies the conditions. Then a₉₉ = 1, a₁₀₀ = -1, so |a₉₉| = |a₁₀₀|. So equality holds.Another example: let a₁ = 2, a₂ = -1, a₃ = 2, a₄ = -1, ..., a₉₉ = 2, a₁₀₀ = -1. Then S₁ = 2 ≥ 0, S₂ = 1 ≤ 0? No, 1 is not ≤ 0. So this doesn't work. So perhaps making the even terms more negative. Let a₁ = 1, a₂ = -1, a₃ = 1, a₄ = -2, etc. But this complicates things.Alternatively, consider a sequence where a₁ is large, and the rest of the terms compensate. For example, let a₁ = 100, and a₂ = -100, then a₃ = 100, a₄ = -100, etc. Then S₁ = 100 ≥ 0, S₂ = 0 ≤ 0, S₃ = 100 ≥ 0, S₄ = 0 ≤ 0, etc. This satisfies the conditions. Here, a₉₉ = 100, a₁₀₀ = -100, so |a₉₉| = |a₁₀₀|. Again equality.But according to my previous reasoning, |a₁₀₀| ≥ |a₉₉|, but in these examples, they are equal. So perhaps the inequality is actually equality, but the problem states |a₉₉| ≥ |a₁₀₀|. Maybe the problem allows for equality, but in the examples, it's possible to have equality. Therefore, maybe the correct answer is |a₉₉| ≥ |a₁₀₀|, and in some cases, equality holds. So why does my initial reasoning suggest the opposite?Let me go back. The key step was:From S₉₉ ≤ a₉₉ and S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉.Since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉. Therefore, a₁₀₀ ≤ -S₉₉ ≤ -a₉₉. Hence, a₁₀₀ ≤ -a₉₉.Thus, since a₁₀₀ ≤ -a₉₉, then -a₁₀₀ ≥ a₉₉. But since a₉₉ ≥ 0, this is |a₁₀₀| ≥ a₉₉ = |a₉₉|. Therefore, |a₁₀₀| ≥ |a₉₉|.But in the examples, |a₁₀₀| = |a₉₉|. So equality holds. Therefore, perhaps the correct inequality is |a₁₀₀| ≥ |a₉₉|, but the problem states the opposite. This suggests either the problem is incorrect, or my reasoning is missing something.Wait, perhaps there's a mistake in the step where S₉₉ ≤ a₉₉. Let me verify that.We have S₉₉ = S₉₈ + a₉₉ ≥ 0.And S₉₈ ≤ 0 (since 98 is even). So S₉₉ = S₉₈ + a₉₉ ≥ 0 ⇒ a₉₉ ≥ -S₉₈.But we also have S₉₈ = S₉₇ + a₉₈. Since S₉₇ ≥ 0 (97 is odd), and a₉₈ ≤ -S₉₇ (from S₉₈ = S₉₇ + a₉₈ ≤ 0 ⇒ a₉₈ ≤ -S₉₇).But S₉₇ ≥ 0 ⇒ -S₉₇ ≤ 0 ⇒ a₉₈ ≤ 0.So a₉₈ is ≤ 0, and similarly for other even terms.But how does this affect S₉₈? S₉₈ is the sum of 98 terms: 49 pairs of (odd, even) terms. Each odd term a_{2k-1} ≥ 0, each even term a_{2k} ≤ something.But perhaps not helpful.Alternatively, since S₉₉ = S₉₈ + a₉₉ ≥ 0 and S₉₈ ≤ 0, then S₉₉ ≥ 0 and S₉₉ ≤ a₉₉. Therefore, 0 ≤ S₉₉ ≤ a₉₉. Therefore, a₉₉ ≥ S₉₉ ≥ 0. Hence, a₉₉ is non-negative and at least as large as S₉₉.Then S₁₀₀ = S₉₉ + a₁₀₀ ≤ 0 ⇒ a₁₀₀ ≤ -S₉₉.Since S₉₉ ≤ a₉₉, then -S₉₉ ≥ -a₉₉ ⇒ a₁₀₀ ≤ -S₉₉ ≥ -a₉₉.Thus, a₁₀₀ ≤ -a₉₉.Therefore, since a₁₀₀ ≤ -a₉₉ and a₉₉ ≥ 0, this implies that |a₁₀₀| = -a₁₀₀ ≥ a₉₉ = |a₉₉|. So |a₁₀₀| ≥ |a₉₉|. But the problem states the opposite. This suggests a contradiction.However, in the examples I considered earlier, equality holds. So unless there's a condition I haven't considered, the problem might be incorrectly stated. Alternatively, perhaps I missed a step in the reasoning.Wait, perhaps there's another inequality involving a₉₉ and a₁₀₀ from a different partial sum.Consider the sum S₁₀₀ = sum from a₁ to a₁₀₀ ≤ 0.Then S₁₀₁ = S₁₀₀ + a₁ ≥ 0. But a₁ ≥ 0, so S₁₀₀ + a₁ ≥ 0 ⇒ S₁₀₀ ≥ -a₁.But S₁₀₀ ≤ 0, so -a₁ ≤ S₁₀₀ ≤ 0 ⇒ a₁ ≥ -S₁₀₀.But how does this relate to a₉₉ and a₁₀₀?Alternatively, consider the sum from a₂ to a₁₀₀, which is S₁₀₀ - a₁. Since S₁₀₀ ≤ 0 and a₁ ≥ 0, this sum is ≤ -a₁ ≤ 0. But this sum is also the sum from a₂ to a₁₀₀, which is 99 terms. Since 99 is odd, this sum should be ≥ 0. Therefore, we have a contradiction unless the sum is zero.Wait, the sum from a₂ to a₁₀₀ is 99 terms (indices 2 to 100 inclusive). Since 99 is odd, the sum must be ≥ 0. But this sum is S₁₀₀ - a₁ ≤ 0 - a₁ ≤ 0, since a₁ ≥ 0. Therefore, this sum must be both ≥ 0 and ≤ 0. Therefore, it must equal zero. Hence, S₁₀₀ - a₁ = 0 ⇒ S₁₀₀ = a₁.But S₁₀₀ is the sum of the first 100 terms, and since the sequence is periodic, S₁₀₀ = sum of one period = S. Therefore, S = a₁. But we also know that S₁₀₀ ≤ 0, so a₁ = S ≤ 0. But a₁ ≥ 0, hence a₁ = 0. Therefore, S = 0.So this implies that the sum over one period is zero. Hence, S = 0.Therefore, S₁₀₀ = 0.Then S₁₀₁ = S₁₀₀ + a₁ = 0 + 0 = 0 ≥ 0, which satisfies the condition.Similarly, S₁₀₂ = S₁₀₁ + a₂ = 0 + a₂ ≤ 0 ⇒ a₂ ≤ 0.But also, S₂ = a₁ + a₂ = 0 + a₂ ≤ 0 ⇒ a₂ ≤ 0.But from the periodicity, a₂ = a₁₀₂ = a₂, so it's fixed.But how does this affect a₉₉ and a₁₀₀?Since S = 0, sum of one period is zero. Therefore, a₁ + a₂ + ... + a₁₀₀ = 0.But we also know that a₁ = 0.Therefore, 0 + a₂ + ... + a₁₀₀ = 0 ⇒ a₂ + ... + a₁₀₀ = 0.But from the previous step, the sum from a₂ to a₁₀₀ is zero (since S₁₀₀ - a₁ = 0). Therefore, the sum from a₂ to a₁₀₀ is zero.Furthermore, considering other partial sums. For example, S₉₉ = sum from a₁ to a₉₉ = 0 + a₂ + ... + a₉₉. But since the sum from a₂ to a₁₀₀ is zero, then sum from a₂ to a₉₉ = -a₁₀₀. Therefore, S₉₉ = -a₁₀₀.But S₉₉ must be ≥ 0 (since 99 is odd). Therefore, -a₁₀₀ ≥ 0 ⇒ a₁₀₀ ≤ 0, which we already knew.But S₉₉ = -a₁₀₀. Also, from earlier, we had S₉₉ ≤ a₉₉. Therefore, -a₁₀₀ ≤ a₉₉ ⇒ a₉₉ ≥ -a₁₀₀. Since a₁₀₀ ≤ 0, this is a₉₉ ≥ |a₁₀₀|. Therefore, |a₉₉| ≥ |a₁₀₀|, which is the desired result.Ah! Here we go. So the key was realizing that the sum over the entire period S must be zero. Because when we considered the sum from a₂ to a₁₀₀, which is 99 terms (odd), it must be ≥ 0. But we also have that sum = S₁₀₀ - a₁ = S - a₁. But S₁₀₀ ≤ 0 and a₁ ≥ 0, so S - a₁ ≤ -a₁ ≤ 0. Therefore, the sum from a₂ to a₁₀₀ must be both ≥ 0 and ≤ 0, hence zero. Therefore, S - a₁ = 0 ⇒ S = a₁. But since S ≤ 0 and a₁ ≥ 0, this implies S = a₁ = 0. Therefore, the sum of each period is zero, and a₁ = 0.Then, considering S₉₉ = sum from a₁ to a₉₉ = sum from a₂ to a₉₉ (since a₁ = 0). But the sum from a₂ to a₁₀₀ is zero (since S = 0), so sum from a₂ to a₉₉ = -a₁₀₀. Therefore, S₉₉ = -a₁₀₀. But S₉₉ ≥ 0 ⇒ -a₁₀₀ ≥ 0 ⇒ a₁₀₀ ≤ 0. Also, from earlier, S₉₉ ≤ a₉₉. Therefore, -a₁₀₀ ≤ a₉₉ ⇒ a₉₉ ≥ -a₁₀₀. Since a₁₀₀ ≤ 0, -a₁₀₀ ≥ 0. Therefore, a₉₉ ≥ |a₁₀₀|. But a₉₉ is part of the sum S₉₉ = -a₁₀₀. Therefore, a₉₉ ≥ |a₁₀₀| implies |a₉₉| ≥ |a₁₀₀|, since a₉₉ is non-negative (from S₉₉ = sum of first 99 terms ≥ 0 and a₁ = 0, so sum from a₂ to a₉₉ = -a₁₀₀ ≥ 0 ⇒ each term a₂ to a₉₉ can be positive or negative, but the sum is non-negative. However, we already found that a₉₉ ≥ |a₁₀₀|.Wait, but a₉₉ is part of the sum from a₂ to a₉₉, which equals -a₁₀₀. So if the sum of those 98 terms (a₂ to a₉₉) is -a₁₀₀, then individual terms can vary, but we also have from earlier that S₉₉ = sum from a₁ to a₉₉ = -a₁₀₀, and since S₉₉ ≤ a₉₉ (because S₉₈ ≤ 0, and S₉₉ = S₉₈ + a₉₉), then -a₁₀₀ ≤ a₉₉. Therefore, a₉₉ ≥ -a₁₀₀. But a₁₀₀ ≤ 0 ⇒ -a₁₀₀ ≥ 0. Hence, a₉₉ ≥ |a₁₀₀|. But since a₉₉ is part of the sequence, and the sequence is periodic, but we don't have information about the other terms. However, since a₉₉ is a single term in the sum S₉₉ = -a₁₀₀, and S₉₉ = a₉₉ + sum of a₂ to a₉₈. But since a₁ = 0, sum from a₂ to a₉₈ = S₉₈. So S₉₉ = S₉₈ + a₉₉ = -a₁₀₀.But S₉₈ ≤ 0 (since 98 is even), so -a₁₀₀ = S₉₈ + a₉₉ ≤ 0 + a₉₉ ⇒ -a₁₀₀ ≤ a₉₉ ⇒ a₉₉ ≥ -a₁₀₀. But since a₁₀₀ ≤ 0, this is a₉₉ ≥ |a₁₀₀|. Therefore, |a₉₉| = a₉₉ ≥ |a₁₀₀|. Therefore, proved.Ah, so the missing step was realizing that due to the periodicity and the alternating sum conditions, the total sum over the period must be zero, which leads to S₉₉ = -a₁₀₀. Combining this with the earlier inequality S₉₉ ≤ a₉₉ gives a₉₉ ≥ -a₁₀₀ = |a₁₀₀|, hence |a₉₉| ≥ |a₁₀₀|. Therefore, the conclusion follows.Final Answerboxed{|a_{99}| geq |a_{100}|}

✅:Given an infinite periodic sequence (a_1, a_2, ldots, a_n, ldots) with period 100, we know that the partial sums (S_n = a_1 + a_2 + cdots + a_n) are non-negative for odd (n) and non-positive for even (n). We need to prove that (|a_{99}| geq |a_{100}|).1. Periodicity and Partial Sums: - The sequence is periodic with period 100, so (a_{n+100} = a_n) for all (n). - The partial sums alternate in sign: (S_n geq 0) for odd (n) and (S_n leq 0) for even (n).2. Key Observations: - For (n = 99) (odd), (S_{99} geq 0). - For (n = 100) (even), (S_{100} leq 0). - (S_{100} = S_{99} + a_{100} leq 0), implying (a_{100} leq -S_{99}).3. Sum Over One Period: - Consider the sum (S_{100} = a_1 + a_2 + cdots + a_{100}). Since (S_{100} leq 0) and the sequence is periodic, (S_{100}) must be non-positive. - For (n = 101) (odd), (S_{101} = S_{100} + a_1 geq 0). Since (S_{100} leq 0), this implies (a_1 geq -S_{100}).4. Sum from (a_2) to (a_{100}): - The sum from (a_2) to (a_{100}) (99 terms) must be non-negative (since it is an odd partial sum). However, this sum is (S_{100} - a_1). Given (S_{100} leq 0) and (a_1 geq 0), this sum must be zero. Hence, (S_{100} = a_1).5. Implications: - Since (S_{100} = a_1) and (S_{100} leq 0), we have (a_1 = 0). Therefore, (S_{100} = 0). - The sum from (a_2) to (a_{100}) is zero, implying (S_{99} = -a_{100}).6. Bounding (a_{99}): - From the partial sum (S_{99} = S_{98} + a_{99} geq 0) and (S_{98} leq 0), we get (S_{99} leq a_{99}). - Since (S_{99} = -a_{100}), we have (-a_{100} leq a_{99}). Thus, (a_{99} geq -a_{100}).7. Absolute Values: - Since (a_{100} leq 0), (|a_{100}| = -a_{100}). - Therefore, (a_{99} geq |a_{100}|), implying (|a_{99}| geq |a_{100}|).Thus, the final result is (boxed{|a_{99}| geq |a_{100}|}).

🔑:1. Assume for the sake of contradiction that the sum of the first 100 terms of the sequence is negative. That is, suppose: [ a_1 + a_2 + ldots + a_{100} = -varepsilon < 0 ]2. Select an integer ( N ) such that: [ N varepsilon > a_1 ] This choice ensures that the negative contribution of the sum ((N varepsilon)) exceeds the positive contribution of (a_1).3. Consider the sum of the first 100( N ) + 1 terms: [ a_1 + a_2 + ldots + a_{100N+1} ] Using the periodicity property (where ( a_{i+100k} = a_i ) for any integer ( k )), this sum can be expressed as: [ (a_1 + a_2 + ldots + a_{100})N + a_1 ] Substituting the assumed negative sum: [ 0 leq a_1 + a_2 + ldots + a_{100N+1} = (a_1 + a_2 + ldots + a_{100})N + a_1 = (-varepsilon)N + a_1 = -N varepsilon + a_1 ] Given our choice of ( N ) such that ( N varepsilon > a_1 ), the inequality simplifies to: [ 0 leq -N varepsilon + a_1 < 0 ] This is a contradiction because it suggests ( a_1 + a_2 + ldots + a_{100N+1} ) lies strictly between 0 and a negative number, which is impossible.4. Therefore, the assumption must be false, meaning: [ a_1 + a_2 + ldots + a_{100} geq 0 ]5. Given the conditions from the problem: [ a_1 + a_2 + ldots + a_{99} geq 0 quad text{and} quad a_1 + a_2 + ldots + a_{98} leq 0 ]6. By comparing sums of sequential terms, subtract the second condition from the first: [ (a_1 + a_2 + ldots + a_{99}) - (a_1 + a_2 + ldots + a_{98}) geq 0 - (a_1 + a_2 + ldots + a_{98}) ] This simplifies to: [ a_{99} geq -a_{100} ]7. Using the properties of absolute values: [ left| a_{99} right| geq -a_{100} ] Since ( a_{100} leq 0 ), we have: [ -a_{100} = left| a_{100} right| ]8. Conclusion: Therefore, we conclude: [ left| a_{99} right| geq left| a_{100} right| ] [ boxed{left| a_{99} right| geq left| a_{100} right|} ]

🧠:Okay, so I need to prove that in a right triangle, the height h dropped to the hypotenuse is less than or equal to the radius r of the inscribed circle multiplied by (1 + sqrt(2)). Hmm. Let me start by recalling some properties of right triangles, inscribed circles, and altitudes.First, in a right triangle, the altitude to the hypotenuse relates the two legs. If the legs are a and b, and the hypotenuse is c, then the area is (a*b)/2. The altitude h can be found by area = (c*h)/2, so h = (a*b)/c. That's a formula I remember.For the inscribed circle, the radius r is given by the formula r = (a + b - c)/2. Wait, let me verify that. The radius of the inscribed circle in any triangle is area divided by the semiperimeter. So for a right triangle, the semiperimeter s is (a + b + c)/2. The area is (a*b)/2. Therefore, r = area / s = (a*b)/ (a + b + c). Wait, so maybe I was wrong earlier. Let me check that again.Yes, the formula for the inradius is r = (area)/(semiperimeter). So in this case, area is (a*b)/2, and semiperimeter is (a + b + c)/2. Therefore, r = (a*b)/ (a + b + c). So r = (a*b)/(a + b + c). Okay, got that.Now, the problem wants me to relate h and r by h ≤ r(1 + sqrt(2)). So I need to express h and r in terms of the sides a, b, c, then find a relationship between them. Maybe I can express h/r and show that it's less than or equal to 1 + sqrt(2).Given h = (a*b)/c and r = (a*b)/(a + b + c). Therefore, h/r = (a*b)/c divided by (a*b)/(a + b + c) = (a*b)/c * (a + b + c)/(a*b) ) = (a + b + c)/c = 1 + (a + b)/c. So h/r = 1 + (a + b)/c. Therefore, the inequality h ≤ r(1 + sqrt(2)) is equivalent to h/r ≤ 1 + sqrt(2), which is equivalent to 1 + (a + b)/c ≤ 1 + sqrt(2). Therefore, this simplifies to (a + b)/c ≤ sqrt(2). So now the problem reduces to proving that (a + b)/c ≤ sqrt(2) in a right triangle.Wait, but in a right triangle, by the Pythagorean theorem, c = sqrt(a^2 + b^2). Therefore, (a + b)/sqrt(a^2 + b^2) ≤ sqrt(2). Let me check if this inequality holds.Let me square both sides to avoid dealing with the square root. Then the inequality becomes (a + b)^2 / (a^2 + b^2) ≤ 2. Multiplying both sides by (a^2 + b^2), we get (a + b)^2 ≤ 2(a^2 + b^2). Expanding the left side: a^2 + 2ab + b^2 ≤ 2a^2 + 2b^2. Subtract a^2 + b^2 from both sides: 2ab ≤ a^2 + b^2. Which is equivalent to 0 ≤ a^2 - 2ab + b^2 = (a - b)^2. Which is always true because squares are non-negative. So the inequality (a + b)/c ≤ sqrt(2) holds, with equality when (a - b)^2 = 0, i.e., when a = b. Therefore, in the case of an isosceles right triangle, the equality holds.Therefore, h/r = 1 + (a + b)/c ≤ 1 + sqrt(2), so h ≤ r(1 + sqrt(2)). Therefore, the inequality is proven.Wait, let me just check with an example. Suppose we have a 3-4-5 triangle. Then h = (3*4)/5 = 12/5 = 2.4. The semiperimeter is (3 + 4 + 5)/2 = 6. The inradius r = (3*4)/6 = 12/6 = 2. Then r(1 + sqrt(2)) = 2*(1 + 1.414) ≈ 2*2.414 ≈ 4.828. And h is 2.4, which is indeed less than 4.828. If we take an isosceles right triangle, legs of length 1, so hypotenuse sqrt(2). Then h = (1*1)/sqrt(2) = 1/sqrt(2). The semiperimeter is (1 + 1 + sqrt(2))/2 = (2 + sqrt(2))/2. The inradius r = (1*1)/2 divided by semiperimeter? Wait, no, area is (1*1)/2 = 0.5. Semiperimeter s = (1 + 1 + sqrt(2))/2. So r = area / s = 0.5 / [(2 + sqrt(2))/2] = 0.5 * 2 / (2 + sqrt(2)) = 1 / (2 + sqrt(2)). Rationalizing the denominator: multiply numerator and denominator by (2 - sqrt(2)), get [ (2 - sqrt(2)) ] / [ (2 + sqrt(2))(2 - sqrt(2)) ) ] = (2 - sqrt(2))/ (4 - 2) ) = (2 - sqrt(2))/2. So r = (2 - sqrt(2))/2. Then r(1 + sqrt(2)) = (2 - sqrt(2))/2 * (1 + sqrt(2)) = [ (2)(1 + sqrt(2)) - sqrt(2)(1 + sqrt(2)) ] / 2 = [ 2 + 2sqrt(2) - sqrt(2) - 2 ] / 2 = [ (2 - 2) + (2sqrt(2) - sqrt(2)) ] /2 = [ sqrt(2) ] / 2. Wait, but h was 1/sqrt(2) ≈ 0.707. So r(1 + sqrt(2)) = sqrt(2)/2 ≈ 0.707, which equals h. So equality holds here, as expected. So that example checks out.Therefore, the steps seem correct. The key was expressing h/r in terms of (a + b)/c, then showing that (a + b)/c ≤ sqrt(2). Which is proven by expanding the square and showing it's equivalent to (a - b)^2 ≥ 0. Therefore, the inequality holds with equality when a = b. So the original inequality h ≤ r(1 + sqrt(2)) is valid for all right triangles, with equality when the triangle is isosceles right-angled.Final AnswerThe inequality is proven to hold, and thus the solution is boxed{h leqslant r(1+sqrt{2})}.

✅:To prove that the height ( h ) of a right triangle dropped to the hypotenuse and the radius ( r ) of the inscribed circle are related by the inequality ( h leqslant r(1+sqrt{2}) ), we start by recalling the relevant formulas for a right triangle with legs ( a ) and ( b ), hypotenuse ( c ), area ( frac{1}{2}ab ), semiperimeter ( s = frac{a + b + c}{2} ), and inradius ( r ).1. Height to the Hypotenuse: [ h = frac{ab}{c} ]2. Inradius: [ r = frac{text{area}}{text{semiperimeter}} = frac{frac{1}{2}ab}{frac{a + b + c}{2}} = frac{ab}{a + b + c} ]3. Ratio ( frac{h}{r} ): [ frac{h}{r} = frac{frac{ab}{c}}{frac{ab}{a + b + c}} = frac{a + b + c}{c} = 1 + frac{a + b}{c} ]We need to show that ( frac{h}{r} leqslant 1 + sqrt{2} ), which simplifies to: [ 1 + frac{a + b}{c} leqslant 1 + sqrt{2} implies frac{a + b}{c} leqslant sqrt{2} ]4. Using the Pythagorean Theorem: [ c = sqrt{a^2 + b^2} ]5. Squaring Both Sides: [ left( frac{a + b}{c} right)^2 leqslant 2 implies frac{(a + b)^2}{a^2 + b^2} leqslant 2 ]6. Expanding and Simplifying: [ (a + b)^2 leqslant 2(a^2 + b^2) implies a^2 + 2ab + b^2 leqslant 2a^2 + 2b^2 implies 0 leqslant a^2 - 2ab + b^2 implies 0 leqslant (a - b)^2 ]This inequality is always true since the square of a real number is non-negative. Equality holds when ( a = b ), i.e., in an isosceles right triangle.Thus, the inequality ( frac{a + b}{c} leqslant sqrt{2} ) holds, leading to: [ h leqslant r(1 + sqrt{2}) ]The final answer is:[boxed{h leqslant r(1+sqrt{2})}]

🔑:1. Identify the relevant geometric elements: Consider a right triangle with legs ( a ) and ( b ), and hypotenuse ( c ). Let ( h ) be the altitude to the hypotenuse, and let ( r ) be the radius of the inscribed circle. For a right triangle, the radius ( r ) of the inscribed circle can be expressed as: [ r = frac{a + b - c}{2} ]2. Express the altitude ( h ): The altitude ( h ) to the hypotenuse in a right triangle is given by the formula: [ h = frac{ab}{c} ]3. Use the Pythagorean theorem and simplifications: By the Pythagorean theorem for a right triangle: [ c = sqrt{a^2 + b^2} ]4. Relate ( h ) and ( r ): We need to establish a relationship between ( h ) and ( r ). Start by expressing the altitude ( h ) in terms of ( r ): [ h text{ (altitude to hypotenuse)} = frac{ab}{c} ]5. Comparison approach using geometric argument: From the reference solution, we want to compare the length ( h - r ) (the leg of a newly formed right triangle) against ( r sqrt{2} ) (hypotenuse of small right triangle). Here is the detailed analytical step: Consider the coordinates of the points on the right triangle. Place the right triangle in the coordinate system where the right angle is at the origin ((0,0)): - One leg along the ( x )-axis: ( A(a,0) ) - Another leg along the ( y )-axis: ( B(0,b) ) - Hypotenuse from origin to ((a,b)) The altitude from the origin to the hypotenuse drops perpendicularly onto the line segment joining ((a,0)) and ((0,b)). Compare: [ h - r ] and use that [ r sqrt{2} ] Finally: [ h leq r(1 + sqrt{2}) ]6. Conclusion: Establishing the final inequality: [ h leq r(1 + sqrt{2}) ] The geometric argument verifies that the altitude to the hypotenuse ( h ) does indeed satisfy the inequality with radius ( r ) of the inscribed circle. [ boxed{h leq r(1 + sqrt{2})} ]