简单做一下：易验证知p>2,即p≥3.p^4的全体约数之和=1+p+p^2+p^3+p^4=x^2则x^2≡1(modp),由于p≥3,所以x≡1(modp)或者x≡-1(modp)对这两种情形分别讨论：1.x≡-1(modp)设x=kp-1,则x^2=k^2p^2-2kp+1=1+p+p^2+p^3+p^4k...