I am not able to solve this equation. Can anybody help?
$p$, $q$ are prime numbers and $a$ is a positive integer.
$$ \frac{pq}{p+q}=\frac{a^2+1}{a+1} $$
The task is to find ALL possible pairs of $p,q$ for this equation.
I've already rewritten that as:
$$ p \cdot q \cdot a - p \cdot a^2 - q \cdot a^2 = -p \cdot q + p + q $$
and I found that one solution is $\{p=2,q=2,a=1\}$.
We write down a proof for $p$ and $q$ distinct odd primes. The other cases are easier to deal with, using easier variants of the idea used below.
If $a$ is even we have a parity problem, so we can take $a$ odd, say $a=2b+1$. Then $\dfrac{a^2+1}{a+1}=a-1+\dfrac{2}{a+1}= 2b+\dfrac{1}{b+1}$. This is a unit fraction more than an integer.
Divide $pq$ by $p+q$. We get $pq=k(p+q)+r$, where $0\lt r\lt p+q$. So $\dfrac{pq}{p+q}$ is an integer plus a fraction, which must be a unit fraction. If follows that $r$ divides $p+q$. But since $pq=k(p+q)+r$, we conclude that $r$ divides $pq$. So $r=1$.
Thus we must have $b=p+q-1$, and $k=2b=2(p+q-1)$. We have arrived at the equation $$pq=2(p+q-1)(p+q)+1.$$ This is impossible, the right-hand side is greater than the left-hand side.