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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 $\{p=2,q=2,a=1\}$.

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Hint: you might be able to show that the two sides of your equation are in their lowest terms (except in certain possible special cases). That gives you some easier material to work with - the solution you have found reflects a special case. –  Mark Bennet Dec 2 '12 at 19:52
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1 Answer 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.

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