求(x^4+1)(y^4+z^4)=t^2的非平凡的正整数解

wayne

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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northwolves

$(x^4+1)(y^4+z^4)=t^2$
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：$p=\sqrt{\frac{76t}{10001}},q=\sqrt{\frac{65t}{10001}}$

mathe

\(y^4+z^4=\frac{t^2}{a}\)
可以改为
\(1+\left(\frac{z}{y}\right)^4=\frac1a \left(\frac{t}{y^2}\right)^2\)
所以变成这个椭圆曲线有理解问题是吧？

wayne

针对 $(x^4 + 1)*(a^4 + 1) = t^2$， 可以通过$[X,T]\to[-\frac{2 \left(a^4+1\right) \left(a^2 x^2+t+1\right)}{(a-x)^2},-\frac{4 \left(a^4+1\right) \left(\left(a^4+1\right) a x^3+a^4+a^3 t x+t+1\right)}{(a-x)^3}]$
得到椭圆曲线$T^2 =X^3 - 4 (a^4+1)^2 X$,
逆变换是$[x,t]=[-\frac{2 \left(a^4+1\right) \left(2 \left(a^5+a\right)+T\right)+a X^2}{4 \left(a^4+1\right)^2+4 \left(a^4+1\right) a^2 X-X^2},\frac{\left(a^4+1\right) \left(4 a^3 T X^2-16 \left(a^4+1\right)^2 \left(\left(a^4+1\right)^2+a^3 T\right)+4 a^2 \left(a^4+1\right) X^3+16 a \left(a^4+1\right) X \left(a^9+2 a^5+a+T\right)+X^4\right)}{\left(-4 \left(a^4+1\right)^2-4 \left(a^4+1\right) a^2 X+X^2\right)^2}]$