Prime power, prime exponent

Note that $p > 1$ because it's a prime number. We divide into two cases.

Case 1 : $x > 1$

Since $x,p > 1$ , this means $p^x, y^p$ are perfect powers. By Mihăilescu's theorem , the only pair of perfect powers that differ by 1 is $2^3 = 8$ and $3^2 = 9$ . This gives $(p,x,y) = (3,2,2)$ .

Case 2 : $x = 1$

Then $p = y^p + 1$ . If $p$ is odd, then $y^p + 1$ factorizes to $(y+1) (y^{p-1} - y^{p-2} + \ldots - y + 1)$ , which means $y^p + 1$ is not prime. Thus $p$ must be even, giving $p = 2$ . This leads to $y = 1$ , so the solution is $(p,x,y) = (2,1,1)$ .

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Thus the only two solutions are $(p,x,y) = (2,1,1), (3,2,2)$ , giving sum of 11.

Case $p=2$ : If $x>1$ then $2^x\equiv0\mod4$ and since $y$ is odd, its square is then $\equiv1\mod 4$ , so the difference is $0-1\equiv-1\equiv\mod4$ but on the RHS we have $\equiv 1\mod4$ , hence whenever $p=2$ we have $x\le1$ , by trial and error one finds the only solution: $x=1,y=1,p=2$

Case $p>2$ : If $p\mid y$ then $y=k\cdot p$ and hence $p^x-(kp)^p=1$ is never reached. (consider modulo $p$ ), so $p\not\mid y$ and the LTE conditions are satisfied. Then we write the original equation as $y^p+1^p=p^x$ and obtain:

$v_p(y^p+1^p)=v_p(p^x)\iff v_p(p)+v_p(y+1)=x\iff v_p(y+1)=x-1$

We then use the inequality; $p^{v_p(y+1)}\le y+1$ to get

$p^{(x-1)}\le y+1\iff p^x\le(y+1)p$

Finally we arrive at the following inequality;

p(y+1)-y^p\ge 1\tag 1

We now use Fermat's little theorem;

$y^p\equiv y\mod p$ and $y^p+1=p^x$ gives that $p\mid(y+1)$ and since $p>2$ this yields $y\ge 2$ and then $(1)$ cannot hold for large $p$ (this is left to you), namely only for $x=2,y=2,p=3$