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How to Solve this?

How Many Ordered Pairs For $$p,q$$ exist if

$${p}^{2}+7pq+{q}^{2}$$ is the Square of an Integer?

EDIT:- p,q are Reals

Note by Mehul Arora
2 years ago

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Ar you sure you want "p,q are Reals" instead of "p,q are integers"? Staff · 2 years ago

It's infinite. · 2 years ago

How? Proper Solution Please? · 2 years ago

It is infinte, let p=0 · 2 years ago

I did initially thought that there are finite solutions, but after seeing @Pi Han Goh sir's solution I was convinced. · 2 years ago

no actually its finite

there are two pairs 3,11 and 11,3 · 2 years ago

HOW? I SAW THAT IN THE RMO SOLUTION AS WELL! · 2 years ago

there could be infinite.

actually the question was for primes. u r asking for positive integers. · 2 years ago

Oh yes. If the original question was for primes it is finite, what do you say @Vaibhav Prasad ??? · 2 years ago

yes u r correct · 2 years ago

BTW, has it really appeared in RMO? · 2 years ago

Yeah.... · 2 years ago

Yes?? · 2 years ago

RMO 2001 · 2 years ago

To be precise the question has not specified what $$p,q$$ must belong to. · 2 years ago

But this question does not specify $$p,q$$ to be primes. · 2 years ago

I think I have solved this before. Anyway , I have a solution but it works only if $$p,q$$ are prime positive integers. $$\ddot\frown$$ · 2 years ago

Can you tell me your method. $$\ddot \smile$$ · 2 years ago

Well , if you see the official solution of this question in RMO , they have done by completing $$(p+q)^2$$ whereas i did it by completing $$(p-q)^2$$. The rest of the method to get the answer is same but only my method has more cases since i have $$9pq$$ whereas the official solution has $$5pq$$. The advantage of official solution is that $$5$$ is a prime. · 2 years ago

Infinite · 2 years ago

Set $$p=q$$ or set $$p=8q$$ shows that there's infinite number of solutions. · 2 years ago

There is another method sir · 2 years ago

Yes, set $$p=0$$ or $$q= 0$$ or $$q = 8p$$. · 2 years ago

But how you got p = 8q one thing ? · 2 years ago

Bound it: WLOG assume $$p,q>0$$. $$(p+q)^2 = p^2 + 2pq + q^2 < p^2 + 7pq + q^2 < p^2 + 8pq + 16q^2 = (p+4q)^2$$, then $$p^2 + 7pq + q^2 = (p+2q)^2 \text{ or } (p+3q)^2$$. · 2 years ago

Exactly sir. · 2 years ago

See RMO 2001 solution · 2 years ago