Finding all prime numbers $p$ such that $p^a + p^b$ is a perfect square

Question

Find all prime numbers $p$ and positive integers $a$ and $b$ such that $p^a + p^b$ a perfect square.

How can I find this. I have no idea about this problem.

Answer 1

Hint: For a first step, consider only the remainders modulo $4$.

Answer 2

Hint: Take $p^a+p^b=n^2$. Factor left hand side to show that you can assume that $a=0$. Then write $n=m+1$ to get $p^b=m^2+2m$ and follow up on that.

Answer 3

Take first $a=b$. Then we want $2p^a$ to be perfect square, which happens if $p=2$ and $a$ is odd. That gives one infinite family of solutions.

Now without loss of generality we may take $a\lt b$. So we want $p^a(1+p^{b-a})$ to be a perfect square. Thus $a$ must be even and $1+p^{b-a}$ must be a perfect square. Let $p^{b-a}+1=x^2$. Then $p^{b-a}=(x-1)(x+1)$.

If $p$ is odd this forces $p=3$ and $b-a=1$. That gives the family of solutions $p=3$, $a=2t$, $b=2t+1$.

If $p=2$, then $x$ must be $3$, for $3$ is the only $x$ such that $x-1$ and $x+1$ are powers of $2$. That gives the family $p=2$, $a=2t$, $b=2t+3$.