(xy+1)(xy+x+2)=k²,展开成y(y+1)x²+(3y+1)x+2-k²=0
Δ= (3y+1)²-4y(y+1)(2-k²)= (y-1)²+4y(y+1)k²是完全平方数,设为n² (n≥0)
对关于n, k的方程(y-1)²+4y(y+1)k²= n²,如果(n, k)是一组整数解,那((2y+1)n-4y(y+1)k , (2y+1)k-n )也是一组整数解
所以再迭代一次得到(n', k')= ( (8y²+8y+1)n-8y(y+1)(2y+1)k,(8y²+8y+1)k-2(2y+1)n )也是一组解,并且 n' = (8y²+8y+1)n-8y(y+1)(2y+1)k ≡ n (mod 2y(y+1))
如果x= [n-(3y+1)] / 2y(y+1) 是整数,那x'=[(8y²+8y+1)n-8y(y+1)(2y+1)k-(3y+1)] / 2y(y+1) 也是整数
而(y-1)²+4y(y+1)k²= n²的解当|k|≥2时,都有n>3y+1,|k|=1时无解,|k|=0时n=y-1
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任意给定正整数y≥1,假设x是原方程最小正整数解,对应的n≡n'≡3y+1 (mod 2y(y+1)),所以不可能n'=y-1, k'=0,那x'= [n'-(3y+1)] / 2y(y+1) 也是正整数,所以x'≥x,化简得n≥(2y+1)k
说明k≤y-1,否则若y-1<k,n²=(y-1)²+4y(y+1)k²< k²+4y(y+1)k²= (2y+1)²k²,n<(2y+1)k,矛盾
k≤y-1时,可以证明n-(2y+1)k≤y-1,否则n>(2y+1)k+y-1,(y-1)²+4y(y+1)k²=n²> (2y+1)²k²+(y-1)²+2(y-1)(2y+1),可得k²+2(y-1)(2y+1)<0,矛盾
而(n'', k'')= ( (2y+1)n-4y(y+1)k , n-(2y+1)k ) 此时是关于n, k的方程n²=(y-1)²+4y(y+1)k² 一组正整数解,由k'' = n-(2y+1)k≤y-1 对应得到 n'' = (2y+1)n-4y(y+1)k≤(2y+1)(y-1)= 2y²-y-1 ① (因为n''²≤(y-1)²+4y(y+1)(y-1)²=(2y+1)²(y-1)²)
又n'' = (2y+1)n-4y(y+1)k ≡ (2y+1)(3y+1) = 6y²+6y+1-y ≡1-y (mod 2y(y+1)),因为n''>0≥1-y,则n''≥2y²+2y+1-y = 2y²+y+1 ②, 和①矛盾
这说明只能原方程无解