Finding all prime $p$ for which there exists a positive integer $n$ such that $p^n+1$ is a perfect square?

Question

$p=7~(n=1)$ is a solution. But how to prove that there are not any other solutions?

Answer 1

We are trying to solve:

$$p^{n}+1=q^2$$

$$p^{n}=(q-1)(q+1)$$

This means that:

$$ \left\{\begin{matrix} q-1=p^{\beta}\\ q+1=p^{\alpha} \\ \alpha>\beta, \alpha+\beta =n \end{matrix}\right. $$

Subtracting the $2$ equations:

$$p^{\alpha}-p^{\beta}=2$$

$$p^{\beta}(p^{\alpha-\beta}-1)=2$$

Since the two factors on the LHS are both integers we have that $p^{\beta}|2$, that is to say $(p,\beta)=(2,1)$. So $p^{\alpha-\beta}-1=2^{\alpha-1}-1=1$ which implies that $\alpha=2$. So $$(p,q,n)=(2,3,3)$$

:)

Answer 2

$p^n+1=a^2$ than $p^n=a^2-1=(a-1)(a+1)$. If $a-1=1$ than $a=2$ and we have $p=3,n=1, a=2$. If $a>2$ we get that difference between 2 divisors of $p^n$ equal 2. $p^k-p^l=p^l(p^{k-l}-1)=2$. So $p=2, l=1, k=2$, or $p=2, n=3, a=3$