求解丢番图方程1/x^2+1/y^2=1/z^2+1/w^2

A Solution to the Diophantus Equation 1/x^2+1/y^2=1/z^2+1/w^2

利用高斯二平方和定理求解一个特殊的丢番图方程1/x^2+1/y^2=1/z^2+1/w^2,将其转化为a^2+b^2=c^2+d^2.经讨论得知,a^2+b^2≡c^2+d^2≡1,2(mod 4),当(k_1-k_3)(k_1+k_3-1)≡(k_4+k_2)(k_4-k_2)时,a^2+b^2≡c^2+d^2≡1(mod4);当(k_1-k_3)(k_1+k_3-1)≡(k_4-k_2)(k_4+k_2-1)≡0.2(mod 4)时,a^2+b^2≡c^2+d^2≡2(mod 4).

In this article,the sum of two squares and Gauss theorem is used to solve a particular diophantus equation 1/x^2＋1/y^2=1/z^2＋1/w^2.1/x^2＋1/y^2=1/z^2＋1/w^2 will be converted to a^2 ＋b^2 =c^2 ＋d^2.After discussion,a^2＋b^2≡c^2＋d^2≡1,2（mod 4）.We say that a^2＋b^2=c^2＋d^2=1（mod 4） if and only if（k1-k3）（k1＋k3-1）≡（k4＋k2）（k4-k2）O（mod4）;we say that a^2 ＋b^2≡c^2 ＋d^2≡2（mod 4） if and only if（k1-k3）（k1 ＋k3- 1）≡（k4-k2）（k4＋k2-1）≡0,2（mod 4）.