Solve for all $n \in \mathbb{N}$ such that: $$n=x^y=\left(x+y\right)^2$$ where $x,y \in \mathbb{Z+}$, and $x$ or $y$ is a prime number.
this is the question, I only got a triple $(n,x,y)=(64,2,6)$.
Can someone verify my solution to this? If so, I'll post it in the comments.
From $x^y=(x+y)^2$ , we get $x^{y/2}=x+y \implies y=2t$ .
If $y$ is prime, then $y=2$ which gives no solution and $y>2$ because $x^2<(x+y)^2$
So $x$ is prime, therefore if $x\ge5$ we can write $x=5+n$, with $n\ge0$.
$(5+n)^y=(5+n+y)^2$, but $y<3$ because $(5+n)^3>(8+n)^2$ $\forall n\ge0$
thus $3>y>2$ if $x\ge5$
$x=3$ don't give solution and $x=2\implies y=6 $ and $n=64$