Indefinite Equation

We guess there is no such $p$

Use apagoge.

Assume exist prime $p$ and integer $n$ ,s.t. $p^p-(p-1)^{p-1}=n^2$

It is easy to know that $p≡1(mod 4)$

Let $p=x^2+y^2$ , Might as well assume $0<x<y$

∵ $p-1$ is not divisible by $p$ ,by the Gauss Decomposition Theorem,we have

$(x+yi)^p=w[(p-1)^{\frac{p-1}{2}}±ni]$ , $w=±1,±i$

∴ $x+yi≡w(1±ni)(mod p)$

Therefore, $x=1,w=±1$ or $y=p-1,w=±i$

∵ $y^2<p$

∴ only , $x=1,w=±1$ is satisfied.

When $w=±1$ ,

$(x+yi)^p=w[(p-1)^{\frac{p-1}{2}}±ni]≡1≡±ni(mod y)$

∴ $y=1$ ,paradoxical!

So,there are not prime $p$ s.t. $p^p-(p-1)^{p-1}=n^2$ and $n$ is integer.

Q.E.D.