a) Show that if \(x\) and \(y\) are in \(A\), then \(xy\) is in \(A\).

b) Show that 11 is not in \(A\).

c) Show that the equation \(x^2 + 4xy + y^2 = 1\) has infinitely many integer solutions.

No vote yet

1 vote

×

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:

TopNewestc) \(x^2+4xy+y^2=(x+2y)^2-3y^2\) which is of the form \(a^2-3b^2\). Since \(3\) is not a perfect square by Pell's equation there are infinitely many solutions.

Log in to reply

a)

\(n=a^2+4ab+b^2=(a+2b)^2-3b^2\).So \(x\) is of the form \(m^2-3n^2\). Similarly \(y\) is of the form \(p^2-3q^2\).

By Brahmgupta's identity, \(xy=(m^2-3n^2)(p^2-3q^2)=(mp+3pq)^2-3(mq+np)^2\) which is also of the form of \(n\).

Log in to reply

b) All perfect squares are of the form \(0,1 \pmod 4\). So \(x^2+4xy+y^2 \equiv 0,1,2 \pmod 4\) but \(11 \equiv 3\pmod 4\).

So no solution.

Log in to reply