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

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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TopNewestAr you sure you want "p,q are Reals" instead of "p,q are integers"? – Calvin Lin Staff · 2 years ago

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It's infinite. – Sravanth Chebrolu · 2 years ago

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– Mehul Arora · 2 years ago

How? Proper Solution Please?Log in to reply

– Archit Boobna · 2 years ago

It is infinte, let p=0Log in to reply

@Pi Han Goh sir's solution I was convinced. – Sravanth Chebrolu · 2 years ago

I did initially thought that there are finite solutions, but after seeingLog in to reply

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

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– Mehul Arora · 2 years ago

HOW? I SAW THAT IN THE RMO SOLUTION AS WELL!Log in to reply

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

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@Vaibhav Prasad ??? – Sravanth Chebrolu · 2 years ago

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

– Vaibhav Prasad · 2 years ago

yes u r correctLog in to reply

– Sravanth Chebrolu · 2 years ago

BTW, has it really appeared in RMO?Log in to reply

– Harsh Shrivastava · 2 years ago

Yeah....Log in to reply

@Harsh Shrivastava – Vaibhav Prasad · 2 years ago

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– Harsh Shrivastava · 2 years ago

Yes??Log in to reply

– Vaibhav Prasad · 2 years ago

RMO 2001Log in to reply

– Nihar Mahajan · 2 years ago

To be precise the question has not specified what \(p,q\) must belong to.Log in to reply

– Nihar Mahajan · 2 years ago

But this question does not specify \(p,q\) to be primes.Log in to reply

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

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– Rajdeep Dhingra · 2 years ago

Can you tell me your method. \(\ddot \smile\)Log in to reply

– Nihar Mahajan · 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.Log in to reply

Infinite – Rajdeep Dhingra · 2 years ago

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Set \(p=q \) or set \(p=8q\) shows that there's infinite number of solutions. – Pi Han Goh · 2 years ago

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– Rajdeep Dhingra · 2 years ago

There is another method sirLog in to reply

– Pi Han Goh · 2 years ago

Yes, set \(p=0 \) or \(q= 0 \) or \(q = 8p\).Log in to reply

– Rajdeep Dhingra · 2 years ago

But how you got p = 8q one thing ?Log in to reply

– Pi Han Goh · 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 \(p^2 + 7pq + q^2 = (p+2q)^2 \text{ or } (p+3q)^2 \).Log in to reply

– Rajdeep Dhingra · 2 years ago

Exactly sir.Log in to reply

See RMO 2001 solution – Vaibhav Prasad · 2 years ago

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@Nihar Mahajan , @Sravanth Chebrolu , @Parth Lohomi , @Rajdeep Dhingra , @Archit Boobna, @Jon Haussmann Sir. Help! – Mehul Arora · 2 years ago

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