Help Please!

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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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Mehul Arora - 6 years, 1 month ago

See RMO 2001 solution

Vaibhav Prasad - 6 years, 1 month ago

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

Pi Han Goh - 6 years, 1 month ago

There is another method sir

Rajdeep Dhingra - 6 years, 1 month ago

Yes, set $p=0$ or $q= 0$ or $q = 8p$ .

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – Exactly sir.

Rajdeep Dhingra - 6 years, 1 month ago

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

Rajdeep Dhingra - 6 years, 1 month ago

@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 - 6 years, 1 month ago

Infinite

Rajdeep Dhingra - 6 years, 1 month ago

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

Nihar Mahajan - 6 years, 1 month ago

Can you tell me your method. $\ddot \smile$

Rajdeep Dhingra - 6 years, 1 month 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.

Nihar Mahajan - 6 years, 1 month ago

It's infinite.

Sravanth C. - 6 years, 1 month ago

How? Proper Solution Please?

Mehul Arora - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

It is infinte, let p=0

Archit Boobna - 6 years, 1 month ago

no actually its finite

there are two pairs 3,11 and 11,3

Vaibhav Prasad - 6 years, 1 month ago

@Vaibhav Prasad – But this question does not specify $p,q$ to be primes.

Nihar Mahajan - 6 years, 1 month ago

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

Mehul Arora - 6 years, 1 month ago

@Mehul Arora – there could be infinite.

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

Vaibhav Prasad - 6 years, 1 month ago

@Vaibhav Prasad – To be precise the question has not specified what $p,q$ must belong to.

Nihar Mahajan - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

@Sravanth C. – yes u r correct

Vaibhav Prasad - 6 years, 1 month ago

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

Sravanth C. - 6 years, 1 month ago

@Sravanth C. – RMO 2001

Vaibhav Prasad - 6 years, 1 month ago

@Sravanth C. – Yeah....

Harsh Shrivastava - 6 years, 1 month ago

@Harsh Shrivastava – @Harsh Shrivastava

Vaibhav Prasad - 6 years, 1 month ago

@Vaibhav Prasad – Yes??

Harsh Shrivastava - 6 years, 1 month ago

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

Calvin Lin Staff - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$