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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, 4 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, 4 months ago

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It's infinite. Sravanth Chebrolu · 2 years, 4 months ago

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@Sravanth Chebrolu How? Proper Solution Please? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora It is infinte, let p=0 Archit Boobna · 2 years, 4 months ago

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@Mehul Arora 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, 4 months ago

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@Mehul Arora no actually its finite

there are two pairs 3,11 and 11,3 Vaibhav Prasad · 2 years, 4 months ago

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@Vaibhav Prasad HOW? I SAW THAT IN THE RMO SOLUTION AS WELL! Mehul Arora · 2 years, 4 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, 4 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, 4 months ago

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@Sravanth Chebrolu yes u r correct Vaibhav Prasad · 2 years, 4 months ago

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@Vaibhav Prasad BTW, has it really appeared in RMO? Sravanth Chebrolu · 2 years, 4 months ago

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@Sravanth Chebrolu Yeah.... Harsh Shrivastava · 2 years, 4 months ago

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@Harsh Shrivastava @Harsh Shrivastava Vaibhav Prasad · 2 years, 4 months ago

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@Vaibhav Prasad Yes?? Harsh Shrivastava · 2 years, 4 months ago

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@Sravanth Chebrolu RMO 2001 Vaibhav Prasad · 2 years, 4 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, 4 months ago

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@Vaibhav Prasad But this question does not specify \(p,q\) to be primes. Nihar Mahajan · 2 years, 4 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, 4 months ago

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@Nihar Mahajan Can you tell me your method. \(\ddot \smile\) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra 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, 4 months ago

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Infinite Rajdeep Dhingra · 2 years, 4 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, 4 months ago

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@Pi Han Goh There is another method sir Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Yes, set \(p=0 \) or \(q= 0 \) or \(q = 8p\). Pi Han Goh · 2 years, 4 months ago

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@Pi Han Goh But how you got p = 8q one thing ? Rajdeep Dhingra · 2 years, 4 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, 4 months ago

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@Pi Han Goh Exactly sir. Rajdeep Dhingra · 2 years, 4 months ago

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See RMO 2001 solution Vaibhav Prasad · 2 years, 4 months ago

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