首先,易見若有整數(shù)解則有正整數(shù)解,不妨設(shè)x,y > 0.
兩式相加得(p+1)² > p²+p+2 = 2y²+2x² = (y+x)²+(y-x)² ≥ (y+x)²,即p+1 > y+x,于是p ≥ y+x.
兩式相減得p(p-1) = 2y²-2x² = 2(y-x)(y+x),有y-x > 0且p | 2(y-x)(y+x).
由x,y > 0,p ≥ y+x > y-x > 0,故p不整除y-x.
易見p = 2時方程無整數(shù)解,故p不整除2.
而p是質(zhì)數(shù),只有p | y+x.又p ≥ y+x > 0,故 p = y+x.
代回得p-1 = 2(y-x).x = (p+1)/4,代入p+1 = 2x²得(p-7)(p+1) = 0.則p = 7.
可驗(yàn)證p = 7時x = 2,y = 5為整數(shù)解.