Finding non-negative integer solutions to $3^x + 5^y = a^2$

Question

$3^x+5^y=a^2$ ($x, y, a$ are non-negative integers)

Find all pairs $(x, y)$ which satisfy the equation.

I have found the trivial solution $x=1, y=0$, and I have tried with congruences, but it didn't really get me anywhere.

Answer 1

By observing modulo 4 one could easily prove that $x$ is an odd number. If $x=1$, by observing modulo 3 it is easy to see that $y$ is an even number. If $x>1$, then $x\ge 3$. So, $27 \mid a^2-5^y$. Looking at modulo 27, we have $y$ is an even number (check every possibility for $0<y<18$ odd and $0<a<14$). Let $y=2t$, $3^x=a^2-5^{2t}=(a+5^t)(a-5^t)$, clearly one of $a+5^t$ or $a-5^t$ is not divisible by 3, so one of them is 1. It follows that $a=5^t+1$. So $2\cdot 5^t+1=3^x$, $2.5^t=3^x-1$. Clearly $x=1$ generate a solution and $x=0,2,3,4$ does not. Clearly $2\mid 3^1-1$ and $5\mid 3^4-1$. For $x>4$, from Zsigmondy's theorem it follows that there are other prime beside $2$ and $5$ that divides $3^x-1$. Hence, the only non-negative integer solution is $x=1$, $y=0$.