If $p=2$, then $25+144=13^2$. If $p>2$, then $5^p+12^p$ is divisible by $17$ and $v_{17}(5^p+12^p)=1+v_{17}(p)$. Therefore, since $v_{17}(5^p+12^p)$ must be even $1= v_{17}(p)$, that is $p=17$. On the other hand $5^{17}\equiv -1$ mod $3$. Contradiction, because squares can’t be equivalent to $-1$ mod $3$. So, answer is $p=2$.

pixi wrote: If $p=2$, then $25+144=13^2$. If $p>2$, then $5^p+12^p$ is divisible by $17$ and $v_{17}(5^p+12^p)=1+v_{17}(p)$. Therefore, since $v_{17}(5^p+12^p)$ must be even $1= v_{17}(p)$, that is $p=17$. On the other hand $5^{17}\equiv -1$ mod $3$. Contradiction, because squares can’t be equivalent to $-1$ mod $3$. So, answer is $p=2$. nice one! But didn't you overcomplicate it using LTE though :p ?

Taking $mod8$ we have $p=even$ so $p=2$

parmenides51 wrote: Find all prime numbers $p$ such that $5^p + 12^p$ is a perfect square Straightforward