求x^2+1=2y^4的所有正整数解

mathe

定义$a_0=0,a_1=1,a_2=2,...,a_{n+1}=2a_n+a_{n-1}$,容易得出$z_n=a_{2n-1}$，递推还可以得出$a_{2n+1}=a_{n+1}^2+a_n^2$
容易算出$a_n$的2的幂次同n中2的幂次相同，而且$a_{2n-1}=1(mod 4)$
可以设$a_n=2^{h(n)}b(n)$,其中$b(n)$是奇数,于是利用二次互反律，我们可以计算雅克比符号
$({a_{2n+2}}/{a_{2n+1}})=({a_{2n}}/{a_{2n+1}})=(2/{a_{2n+1}})^{h(2n)}({b_{2n}}/{a_{2n+1}})=(2/{a_{2n+1}})^{h(2n)}({a_{2n+1}}/{b_{2n}})=(2/{a_{2n+1}})^{h(2n)}({a_{2n-1}}/{b_{2n}})=(2/{a_{2n+1}})^{h(2n)}({b_{2n}}/{a_{2n-1}})=(2/{a_{2n+1}})^{h(2n)}(2/{a_{2n-1}})^{h(2n)}({a_{2n}}/{a_{2n-1}})$
而$(2/{a_{2n-1}})$取决于$a_{2n-1}(mod 8)$的值，而$a_n$模8是以8为周期的，我们应该可以得出$({a_{2n}}/{a_{2n-1}})$也是以8为周期的
另外$a_{n+k}=a_na_{k+1}+a_{n-1}a_k$,我们可以得出对于给定的k,$({a_{2n+k}}/{a_{2n-1}})=({a_{2n}a_k}/{a_{2n-1}})$也是以8为周期的，我们可以先将这些值算出，应该有用

mathe

$a_n(mod 8)$前几项为0,1,2,5,4,5,6,1,0,1,…

mathe

所有对于奇数n有$(2/{a_n})=(2/n)$

mathe

$({a_{8k+2}}/{a_{8k+1}})=({a_{8k}}/{a_{8k-1}})$
$({a_{8k+4}}/{a_{8k+3}})=-({a_{8k+2}}/{a_{8k+1}})$
$({a_{8k+6}}/{a_{8k+5}})=({a_{8k+4}}/{a_{8k+3}})$
$({a_{8k+8}}/{a_{8k+7}})=-({a_{8k+6}}/{a_{8k+5}})$
由此得出
$ ({a_{8k+2}}/{a_{8k+1}})= ({a_{8k+8}}/{a_{8k+7}})=1, ({a_{8k+4}}/{a_{8k+3}})= ({a_{8k+6}}/{a_{8k+5}})=-1$
于是$({a_{8k+3}}/{b_{8k+4}})=-1,({a_{8k+5}}/{b_{8k+6}})=1$

葡萄糖

偶然找到一篇文章
On Diophantine Equation $x^2 = 2y^4-1$
https://wenku.baidu.com/view/89b9880df12d2af90242e64e.html