试求x,y的方程x²+y²=208(x+y)的所有正整数解

x²+y² = 208(x+y)配方为(x-104)²+(y-104)² = 2·104².
设z = x-104,w = y-104,转化为求z²+w² = 2·104²的整数解.
首先z,w同奇同偶.
但奇数的平方除以8余1,两个奇数的平方和不可能被8整除.
故z,w均为偶数,设z = 2u,w = 2v,转化为求u²+v² = 2·52²的整数解.
同样讨论知u,v均为偶数,设u = 2s,v = 2t,转化为求s²+t² = 2·26²的整数解.
又知s,t均为偶数,设s = 2m,t = 2n,只要求m²+n² = 2·13² = 338的整数解.
可知m,n均为奇数,逐个尝试得m = 7,n = 17 与m = n = 13两组满足m ≤ n的正整数解.
于是z²+w² = 2·104²的整数解有12组:
(56,136),(56,-136),(-56,136),(-56,-136),
(104,104),(104,-104),(-104,104),(-104,-104),
(136,56),(136,-56),(-136,56),(-136,-56).
对应原方程的正整数解有:(160,240),(48,240),(208,208),(240,160),(240,48).