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

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## Comments

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TopNewest@Nihar Mahajan , @Sravanth Chebrolu , @Parth Lohomi , @Rajdeep Dhingra , @Archit Boobna, @Jon Haussmann Sir. Help!

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See RMO 2001 solution

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Set \(p=q \) or set \(p=8q\) shows that there's infinite number of solutions.

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There is another method sir

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Yes, set \(p=0 \) or \(q= 0 \) or \(q = 8p\).

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Infinite

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I think I have solved this before. Anyway , I have a solution but it works only if \(p,q\) are prime positive integers. \(\ddot\frown\)

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Can you tell me your method. \(\ddot \smile\)

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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.

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It's infinite.

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How? Proper Solution Please?

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

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It is infinte, let p=0

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no actually its finite

there are two pairs 3,11 and 11,3

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actually the question was for primes. u r asking for positive integers.

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@Vaibhav Prasad ???

Oh yes. If the original question was for primes it is finite, what do you sayLog in to reply

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@Harsh Shrivastava

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Ar you sure you want "p,q are Reals" instead of "p,q are integers"?

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