Discuss a set of three numbers $x, y, z ∈ N$ such that $x^2+5^4=5^y+z$ .
What about the possible pairs of numbers $x, y ∈ N$, $y$ being an even number, such that $x^2+5^4=5^y$ ? What if $y$ being odd? What if we have another prime number instead of $5$?
About the first question, I have found that for $x=0, y=4, z=0$ and $x=50, y=5, z=0$ the equation meet its conditions. I know it is possibile to find even more solutions, but I can't really find a proper way to "discuss" all of them.
About the second one, if $y$ is an even number, the solution I've found is $x=0, y=4$, while if $y$ is an odd number it's $x=50, y=5$.
About the third question, it seems that for every prime number the only thing we can say for sure is that $y≥4$.
Consider $x^2+5^4 = 5^y+z \implies z = x^2-5^y+5^4$ sub to $x,y,z \in N$. There is one equation and three unknowns, which means degree of freedom is two (possibly infinite solutions). Choose any two variables and the third one will be constrained by the equation. Since $x, y$ are appear in square and exponential terms, choose these two first. It makes easy to get the value of z. Since $z \in N$ choose $x, y$ such that $z > 0 \implies 5^y < x^2 + 5^4$. The procedure is
Consider $x^2+5^4=5^y \implies x^2 = 5^y-5^4 = 5^4(5^{y-4}-1) = 125^2 (5^{y-4}-1)$. Things get tricky here, because we have to choose y s.t $z = 5^{y-4}-1$ is a perfect square. Just like you mentioned, when $y=0$ or $y=1$, $z$ is a perfect square. If $y$ is even, then $y-4$ is also even, $(5^{y-4}-1)$ can be written as $(a^2-1)$ where $a = 5^{\frac{y-4}{2}} $. $a^2-1$ will be a perfect square only when $y=0$. Therefore $y=0$ is the only even solution. I'm not sure about the no of solutions, when $y$ is odd. Regarding your third question $y$ should be greater than or equal to 4, because if $ y < 4 \implies 5^{y-4}-1 < 0$. Therefore $y \geq 4$ for any number, not just 5.