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

Note by Swapnil Das 3 years, 3 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.

\(m+p+q=2p+2q+5\). 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;

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## Comments

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TopNewestset \(p^2+7pq+q^2=m^2\), 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.

\(m+p+q=2p+2q+5\). 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;

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Did you mean p=/=q?

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