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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:

$p^2 = -1(textmod , q)$ and $q = -1(textmod , p)$;
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

$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$

Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to

$$p^2-k^2p+(k+1)=0,$$

which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?

Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, 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 $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.

Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?

Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.

From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$

1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$

Since $p^2+1geq s$ we have 2 subcases:

1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$

So $q+1leq p-1 leq q-1$ and thus no solution.

1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.

2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$

so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.