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# RMO doubt

Find all primes $$p$$ and $$q$$ such that $${ p }^{ 2 }+7pq+{ q }^{ 2 }$$ is a perfect square.

Note by Swapnil Das
2 years, 5 months ago

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set $$p^2+7pq+q^2=m^2$$, so

$$(p+q)^2+5pq=m^2$$. So $$5pq= (m+p+q)(m-p-q)$$. It follows, given that $$m-p-q<m+p+q$$, that

$$m-p-q$$ is either $$5, p, q, 5p, or 5q$$. We can assume $$q>p$$ and exclude the last possibility.

• If $$m-p-q=5$$, we get $$m+p+q= pq$$ and

$$m+p+q=2p+2q+5$$. So $$pq=2p+2q+5$$ and $$(p-2)(q-2)=9$$. So the only solution is p=3 and q=11 so m=19.

• If $$m-p-q= p$$, then $$m+p+q=5q$$. Then $$m= 2p+q=4q-p$$, which implies $$p=q$$. This is excluded.

• Same for $$m-p-q=q$$.

• If m-p-q=5p. Then m+p+q = q which is impossible.

So (3,11) and (11,3) are the only solutions;

- 2 years, 5 months ago

Did you mean p=/=q?

- 2 years, 5 months ago