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攒拳怒目的坚果
9月前
证：
$易知只能是x^2+y=p,x+y^2=p^4 或x^2+y=p^2,x+y^2=p^3$
$(1)x^2+y=p,x+y^2=p^4,则x≤\sqrt{p},y≤p,p^4=x+y^2≤x+\sqrt{p}+p^2\lt p^4 矛盾.$
$(2)x^2+y=p^2,x+y^2=p^3,则y\equiv -x^2(mod p^2),0\equiv x+y^2 \equiv x+x^4 (mod p^2)$
$故p^2|x(x+1)(x^2-x+1)$
$(i) p|x,则p≤x,故p^2≤x^2\lt x^2+y=p^2 矛盾.$
$(ii) p^2|x+1,则p^2≤x+1,故p≤p^2-1≤x,与(i)同理矛盾$
$(iii) p^2|x^2-x+1,则p^2≤x^2-x+1≤x^2,故p≤x,与(i)同理矛盾$
$(iv) p|x+1且p|x^2-x+1,则x\equiv -1(mod p),0\equiv x^2-x+1\equiv (-1)^2-(-1)+1=3(mod p)$
$故p=3,讨论x=1与x=2即可知x=2,y=5,则原题所有解为(x,y)=(2,5)或(5,2)$