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※ 引述《starshiny (璀璨星辰)》之銘言: : 試問滿足|a+b|+c=19且ab+|c|=97的整數解(a,b,c)共有幾組? : 自己有嘗試著求解 : 可是怎麼算都只能求出兩組解 : 跟解答差有點多QAQQQQQ : 懇求各位大大一起集思廣益 c <= 19, ab <= 97 a+b = +-(19-c) ab = 97-|c| a, b are roots of x^2 -+ (19-c) + (97-|c|) D = (19-c)^2 - 4(97-|c|) | = c^2 - 34c - 27, c>=0 | = k^2, k integer >= 0 | | | = c^2 - 42c - 27, c <0 | c>=0, (c-17)^2 - k^2 = (c-17-k)(c-17+k) = 289+27 = 316 = 4 * 79 2* 1 2*79 k=78, c-17= 80, c不合 (c <= 19) -2*79 -2* 1 k=78, c-17=-80, c不合 c <0, (c-21)^2 - k^2 = (c-21-k)(c-21+k) = 441+27 = 468 = 4 *117 (c-21-k < 0) -2*117 -2* 1 k=116, c-21=-118, c=-97 -2* 39 -2* 3 k= 36, c-21= -42, c=-21 -2* 13 -2* 9 k= 4, c-21= -22, c= -1 a, b=(1/2)( +-(19-c) +-k ) k | c | 19-c+k | 19-c-k | (a, b) --------------------------------------------------------------------- 116 | -97 | 232 | 0 | (116, 0) (-116, 0) (0, 116) (0, -116) --------------------------------------------------------------------- 36 | -21 | 76 | 4 | (38, 2) (-38, -2) (2, 38) (-2, -38) --------------------------------------------------------------------- 4 | -1 | 24 | 16 | (12, 8) (-12, -8) (8, 12) (-8, -12) 以上騙P幣ow o 其實我原本忘記c <= 19, 驗算的時候多了一組 -- 嗯嗯ow o -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 49.214.167.64 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1475158574.A.17B.html
LPH66 : 啊, 忘記拆 117 了所以只做出四組... 09/29 22:30
starshiny : 看不太懂怎麼變成a,b are roots of x^2 這段 09/29 23:22
Desperato : 根與係數的關係 09/29 23:24
starshiny : 因為沒化成x^2-(a+b)x+ab=0形式一下子沒反應過來XD 09/29 23:27
LPH66 : 其實判別式本質上就是 (a-b)^2 09/30 00:45
LPH66 : 所以直接算這東西也會變成同一個式子 09/30 00:45