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[讨论] 求(x^4+1)(y^4+z^4)=t^2的非平凡的正整数解

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发表于 昨天 15:49 | 显示全部楼层 |阅读模式

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知乎上有一道题,   求$(x^4+1)(y^4+z^4)=t^2$的正整数解. 咱只考虑非平凡解.
我给出了初步的解答, 我发现值得讨论.就暂时先不公布链接了.

比如x=10的时候,有这些非平凡解
  1. (10, 3470, 30101)
  2. (10, 18310, 58019)
  3. (10, 300059999, 999399970)
  4. (10, 132211580450, 1560535879379)
  5. (10, 5778887979110, 11136239888261)
  6. (10, 523606833336281, 1582146687563950)
  7. (10, 3130830486676759, 5854408448039230)
  8. (10, 5556728420267245053079, 42561428523400632386470)
  9. (10, 1609802938066867151963801, 2141085960990531655699990)
  10. (10, 5006198949699987499500001, 9949987496999895006200050)
  11. (10, 2303868130334468166093591370, 8086395583987023071375221541)
  12. (10, 155257989527532997069088980270, 211405025057472201141884259059)
  13. (10, 796115258361316260299119503787370, 1445620056911436957284861910406259)
  14. (10, 7837324310880690090986184344920450, 10184771542621891067453374769662181)
  15. (10, 7943975435658355615444415692551860839, 114393799419193259636813763067876093510)
  16. (10, 7018615996813179744376653975993845738410, 20265991066696150806094057972329533770901)
  17. (10, 76517778735386331358097862503085145042390, 76898412155447329577757461097562989046039)
  18. (10, 7004028079548015956494146485192474788359779, 7192648283396860342485284356535027357722130)
  19. (10, 7030770440144476077248079783356744756130201959999, 9803869652424331217919242267523298522953969199930)
  20. (10, 2035293386112283958905238419846414687850220089944010, 4175769588593750634106517154910239291825434081419139)
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发表于 昨天 22:33 | 显示全部楼层
本帖最后由 northwolves 于 2025-7-16 22:42 编辑

$(x^4+1)(y^4+z^4)=t^2$
以x=10为例,此时$r=(x^4+1)=10001;(\frac{ry^2}{t})^2+(\frac{rz^2}{t})^2=r=10001$
$76^2+65^2=10001$
$p^2+q^2=10001$的一个通解为:$(\frac{76 k^2-130 k-76}{k^2+1},\frac{65 k^2+152 k-65}{k^2+1})$



解得:$p=\sqrt{\frac{76t}{10001}},q=\sqrt{\frac{65t}{10001}}$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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