Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Romania

Find all positive integers $r$ with the property that there exists positive prime numbers $p$ and $q$ so that $p^2 + pq + q^2 = r^2$ .

Solution

The given relation is equivalent to $(p+q)^2 = r^2 + pq$ , which can be written $(p+q+r)(p+q-r) = pq$ .
The divisors of $pq$ are $1$ , $p$ , $q$ and $pq$ . Since $p+q > \max\{p,q\}$ , it follows that $p+q-r = 1$ and $p+q+r = pq$ .

Adding the last two equalities yields $2p+2q = pq+1$ , that is $(p-2)(q-2) = 3$ .
This leads to $(p,q) \in \{(3,5), (5,3)\}$ ; in both cases $r=7$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.