Find the values of $p$ and $q$ 
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers, 
  Find all the solutions (p, q) 

I tried to solve this exercise using that:
$p^2 = -1(\text{mod} \, q)$ and $q = -1(\text{mod} \, p)$; 
   So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers. 
Then I tried to solve a quadratic equation, but I could not finish the problem

Clearly $p\neq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $p\mid q+1$ and $q\mid p^2+1$. Write
$$q+1=ap\qquad\text{ and }\qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1\equiv0\pmod{p}$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
and as $p\geq2$ clearly we cannot have $a,c\geq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1\mid p^2+1$. As $p-1\mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
