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I have the relation $x^2 + 4xy + y^2 = 1$. What I need to do is prove that it has infinitely many integer solutions.

I started out by solving for $y$ and getting $y = -2x \pm \sqrt{3x^2 + 1}$ and this shows that there are infinently many solutions as $x$'s domain is all reals, but I don't know how to prove that it has infinently many integer solutions. Any help would be awesome.

Thanks!

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2 Answers 2

up vote 5 down vote accepted

Completing the square, we get $$x^2+4xy+y^2=x^2+4xy^2+4y^2-3y^2=(x+2y)^2-3y^2=1.$$ Letting $z=x+2y$ (so $x=z-2y$), you now just have to show that there are infinitely many integer solutions to $$z^2-3y^2=1.$$ This is an instance of the famous Pell's equation (Wikipedia).

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We continue from the place you reached. We want to show that there are infinitely many integers $x$ such that $3x^2+1$ is a perfect square, that is, that the equation $w^2=3x^2+1$ has infinitely many integer solutions. It is common to rewrite this as $w^2-3x^2=1$.

This is a Pell equation, and it is a standard fact that it has infinitely many solutions. We give a quick proof.

It is clear that one solution is $w_1=2$, $x_1=1$. One can verify that if $(w_n,x_n)$ is a solution, so is $(w_{n+1},x_{n+1})$, where $$w_{n+1}=w_n +3x_n\quad\text{and}\quad x_{n+1}=w_n+2x_n.$$ This fact is perhaps not easy to find, but it is easy to verify. For $$w_{n+1}^2-3x_{n+1}^2 =(2w_n+3x_n)^2 -3(w_n+2x_n)^2= w_n^2-3x_n^2=1.$$ The pairs $(w_n,x_n)$ give us infinitely many positive integer solutions of the Pellian. It turns out these are all the solutions in positive integers.

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