Is 2 a quadratic residue modulo $r=\frac{p^m+1}{2}$? Suppose that $$r=\frac{p^m+1}{2}$$ is a prime number, where $p$ is also prime. Does the equation $$p^{2}-2\equiv 0 \pmod{r}$$ have any solutions? 
Thanks in advance.
 A: For any prime $\,q\,$, the equation $\,x^2=2\pmod q\,$ has a solution iff $\,q=\pm 1\mod 8\,$ . As I am assuming that in "$\,p^2-2=0\pmod r\,$" you actually  meant $\,x^2-2=0\pmod r\,$ (if I'm wrong in my assumption please disregard this post), I'll give you two examples with different outcomes: 
== for $\,\displaystyle{\,p=3\,,\,m=4\,,\,r=\frac{3^4+1}{2}=41=1\pmod 8\,\,}$ , so your equation has solution
== for $\,\,\displaystyle{p=5\,,\,m=2\,\,,\,r=\frac{5^2+1}{2}=13\neq \pm1\pmod 8}\,\,$ , and your equation has no solution here.
Thus I believe you must either change your question or impose further conditions.
A: I'm going to assume the question as posed is the question intended. 
$p^2-2\equiv0\pmod r$ requires $r\lt p^2$. But $r=(p^m+1)/2$ almost requires $r\gt p^2$, so the two relations are nearly mutually exclusive. The only ways they can both hold are 


*

*$m=1$. Then $r=(p+1)/2$; $p=2r-1$; $p^2-2=4r^2-4r-1\equiv-1\pmod r$, so no solutions here. 

*$m=2$. Then $r=(p^2+1)/2$; $p^2=2r-1$; $p^2-2=2r-3\equiv-3\pmod r$; so the only way this is going to work is if $-3\equiv0\pmod r$, which says $r=3$, but this leads to $p=\pm\sqrt5$, which is nonsense, so no solutions here, either. 
And that's it. If $m\ge3$ then you'll find that $r=(p^m+1)/2\gt p^2$, so there are no other cases to look at. 
In summary, with the problem as posed, the congruence has no solutions. 
A: If you are asking whether $2$ is a quadratic residue of $r$, it is easy to find by treating each of the following 8 cases: when $p \equiv 1, 3, 5, 7, 9, 11, 13, 15 (\bmod\, 16)$ respectively.
Suppose $r=(p^m+1)/2$ is a prime, then the condition for $2$ being a quadratic residue of $r$ is
when $p=16n+1$, $m$ can be any natural number
when $p=16n+3$ or $16n+11$, $m\equiv 0\, (\bmod\, 4)$
when $p=16n+5$, $m\equiv 0\ \mbox{or}\ 3\, (\bmod\, 4)$
when $p=16n+7$ or $16n+9$ or $16n+15$, $m$ can be any even number
when $p=16n+13$, $m\equiv 0\ \mbox{or}\ 1\, (\bmod\, 4)$
The principle is to raise the least residues to (at most) power $4$, plus $1$ then divide by $2$, and find those which $\equiv \pm 1\, (\bmod\, 8)$
