# Algebra

Let $$A$$ be the set of all integers $$n$$ of the form $$n= a^2 + 4ab+ b^2$$ where $$a$$ and $$b$$ are integers.
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.

Note by Bulbuul Dev
1 year, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> 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"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

c) $$x^2+4xy+y^2=(x+2y)^2-3y^2$$ 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.

- 1 year, 5 months ago

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 $$n$$.

- 1 year, 5 months ago

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.

- 1 year, 5 months ago