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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, 7 months ago

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

Calvin Lin Staff - 2 years, 7 months ago

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It's infinite.

Sravanth Chebrolu - 2 years, 7 months ago

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How? Proper Solution Please?

Mehul Arora - 2 years, 7 months ago

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It is infinte, let p=0

Archit Boobna - 2 years, 7 months ago

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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.

Sravanth Chebrolu - 2 years, 7 months ago

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no actually its finite

there are two pairs 3,11 and 11,3

Vaibhav Prasad - 2 years, 7 months ago

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@Vaibhav Prasad HOW? I SAW THAT IN THE RMO SOLUTION AS WELL!

Mehul Arora - 2 years, 7 months ago

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@Mehul Arora there could be infinite.

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

Vaibhav Prasad - 2 years, 7 months ago

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

Sravanth Chebrolu - 2 years, 7 months ago

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@Sravanth Chebrolu yes u r correct

Vaibhav Prasad - 2 years, 7 months ago

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@Vaibhav Prasad BTW, has it really appeared in RMO?

Sravanth Chebrolu - 2 years, 7 months ago

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@Sravanth Chebrolu Yeah....

Harsh Shrivastava - 2 years, 7 months ago

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@Harsh Shrivastava @Harsh Shrivastava

Vaibhav Prasad - 2 years, 7 months ago

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@Vaibhav Prasad Yes??

Harsh Shrivastava - 2 years, 7 months ago

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@Sravanth Chebrolu RMO 2001

Vaibhav Prasad - 2 years, 7 months ago

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@Vaibhav Prasad To be precise the question has not specified what \(p,q\) must belong to.

Nihar Mahajan - 2 years, 7 months ago

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@Vaibhav Prasad But this question does not specify \(p,q\) to be primes.

Nihar Mahajan - 2 years, 7 months ago

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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. \(\ddot\frown\)

Nihar Mahajan - 2 years, 7 months ago

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

Rajdeep Dhingra - 2 years, 7 months ago

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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.

Nihar Mahajan - 2 years, 7 months ago

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Infinite

Rajdeep Dhingra - 2 years, 7 months 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, 7 months ago

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There is another method sir

Rajdeep Dhingra - 2 years, 7 months ago

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

Pi Han Goh - 2 years, 7 months ago

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@Pi Han Goh But how you got p = 8q one thing ?

Rajdeep Dhingra - 2 years, 7 months ago

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@Rajdeep Dhingra 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 \(p^2 + 7pq + q^2 = (p+2q)^2 \text{ or } (p+3q)^2 \).

Pi Han Goh - 2 years, 7 months ago

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@Pi Han Goh Exactly sir.

Rajdeep Dhingra - 2 years, 7 months ago

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See RMO 2001 solution

Vaibhav Prasad - 2 years, 7 months ago

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