So my question is the title; I have been self studying for an exam coming up and this problem showed up. I am not sure how to proceed I can't seem to see how the $n_p(G)$ is related to $p+q$. Any hint is appreciated! Also, I have already finished the case when $p=q$.
EDIT: If $p=q$ then $G$ is of order $2p^2$. Now, if $p=2$, then $|G|=8$. Up to isomorphism we have 5 groups of order $8$ all of which are not simple. So we may suppose that $p$ is an odd prime. In this case every Sylow $p$-subgroup has index two in $G$ and thus is normal. So we may suppose that $q<p$....
Assume that $p > q$. Then we have that $\gcd(p,p+q)=\gcd(p,q) = 1$
Hence by Sylow Theorems we have that $n_p(G) \equiv 1 \pmod p$. Also $n_p(G) \mid p+q$. But now $n_p(G) \le p+q < 2p < 2p+1$. Moreover $p+1 \not \mid p+q$, as $p+2 < p+q < 2p+2$. Therefore we must have that $n_p(G) = 1$. Hence the proof.
As $n_p(G) \equiv 1 \pmod p$ we must have that $n_p(G) = kp+1$ for some non-negative integer $k$. From above we have that $kp+1 = n_p(G) < 2p+1$ and so we have that $k<2$. The case $k=1$ is impossible, as $p+1 \not \mid p+q$ and hence we must have $k=0$. Hence the proof.