how will i find all values of p and q when p and q are prime and p^2+7pq+q^2 is a perfect square?

Question

I mean
p not equal to q and p^2+7pq+q^2 is a square of a natural number.

Answer 1

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 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.
- Idem 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;

Answer 2

gianlino's method has given the answers, but there are some other possibilities we need to look at before we are sure there are no other answers. He observed that (m-p-q) must be a factor of 5pq, and he looked at the possibilities that (m-p-q) is 5, p, q, 5p, or 5q. But there are other possibilities, namely, 1, pq, and 5pq.

If m-p-q = 1 then m+p+q = 5pq,
Also m = 1+p+q, so
5pq = 1 + 2p + 2q.
But this is impossible because
5pq = pq + 2pq + 2pq > 1 + 2p + 2q.

If m-p-q = pq then m+p+q = 5
Also m = pq + p + q, so
pq + 2p + 2q = 5, which is impossible because
pq + 2p + 2q > 1 + 2 + 2 = 5

If m-p-q = 5pq then m+p+q = 1, which is impossible.

So the other possible factors don't yield any new solutions. gianlino's (3,11) and (11,3) are the only ones.

Answer 3

NO u can't get perfect square, because

p^2+2pq+q^2 = (p+q)^2 which is perfect square but not ur question

Answer 4

I agree with the above answer.