RMO doubt

Find all primes pp and qq such that p2+7pq+q2{ p }^{ 2 }+7pq+{ q }^{ 2 } is a perfect square.

#NumberTheory

Note by Swapnil Das
5 years, 10 months ago

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Comments

set p2+7pq+q2=m2p^2+7pq+q^2=m^2, so

(p+q)2+5pq=m2(p+q)^2+5pq=m^2. So 5pq=(m+p+q)(m−p−q)5pq= (m+p+q)(m-p-q). It follows, given that m−p−q<m+p+qm-p-q<m+p+q, that

m−p−qm-p-q is either 5,p,q,5p,or5q5, p, q, 5p, or 5q. We can assume q>pq>p and exclude the last possibility.

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

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

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

  • Same for m−p−q=qm-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;

Aditya Kumar - 5 years, 10 months ago

Did you mean p=/=q?

Gian Sanjaya - 5 years, 10 months ago
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