New user? Sign up

Existing user? Log in

Find all primes \(p\) and \(q\) such that \({ p }^{ 2 }+7pq+{ q }^{ 2 }\) is a perfect square.

Note by Swapnil Das 2 years, 9 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

set \(p^2+7pq+q^2=m^2\), so

\((p+q)^2+5pq=m^2\). So \(5pq= (m+p+q)(m-p-q)\). It follows, given that \(m-p-q<m+p+q\), that

\(m-p-q\) is either \(5, p, q, 5p, or 5q\). We can assume \(q>p\) and exclude the last possibility.

\(m+p+q=2p+2q+5\). So \(pq=2p+2q+5\) and \((p-2)(q-2)=9\). So the only solution is p=3 and q=11 so m=19.

If \(m-p-q= p\), then \(m+p+q=5q\). Then \(m= 2p+q=4q-p\), which implies \(p=q\). This is excluded.

Same for \(m-p-q=q\).

If m-p-q=5p. Then m+p+q = q which is impossible.

So (3,11) and (11,3) are the only solutions;

Log in to reply

Did you mean p=/=q?

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestset \(p^2+7pq+q^2=m^2\), so

\((p+q)^2+5pq=m^2\). So \(5pq= (m+p+q)(m-p-q)\). It follows, given that \(m-p-q<m+p+q\), that

\(m-p-q\) is either \(5, p, q, 5p, or 5q\). We can assume \(q>p\) and exclude the last possibility.

\(m+p+q=2p+2q+5\). So \(pq=2p+2q+5\) and \((p-2)(q-2)=9\). So the only solution is p=3 and q=11 so m=19.

If \(m-p-q= p\), then \(m+p+q=5q\). Then \(m= 2p+q=4q-p\), which implies \(p=q\). This is excluded.

Same for \(m-p-q=q\).

If m-p-q=5p. Then m+p+q = q which is impossible.

So (3,11) and (11,3) are the only solutions;

Log in to reply

Did you mean p=/=q?

Log in to reply