Proof that $x^2+4xy+y^2=1$ has infinitely many integer solutions

The question would, naturally, be very straight forward if there was a $2xy$ instead of a $4xy$. Then it would simply be a matter of doing:

$$
x^2+2xy+y^2=1\\
(x+y)^2=1\\
\sqrt{(x+y)^2}=\sqrt{1}\\
|x+y|=1
$$

Clearly, there are infinitely many pairs of integers that differ by one so there is an infinite number of integer solutions to $x^2+2xy+y^2=1$. Unfortunately, the same principle does not apply to $x^2+4xy+y^2=1$ where if I attempt to construct a similar proof all I can do is:

$$
x^2+4xy+y^2=1\\
(x+y)^2+2xy=1\\
2xy=1-(x+y)^2\\
2xy=(1+x+y)(1-x-y)
$$

And, from there, I have no idea how to proceed to complete the proof that there are an infinite number of integer solutions. I’m wondering whether I’m approaching the question entirely in the wrong way or if I am simply missing something. A nudge in the right direction would be much appreciated!

Solutions Collecting From Web of "Proof that $x^2+4xy+y^2=1$ has infinitely many integer solutions"

Hint $\ $ Consider $\,f(x) = x^2 + 4y\ x + y^2\!-\!1\,$ as a quadratic in $\,x,$ where $y$ is constant. By Vieta its roots $\,x,x’$ satisfy $\ x+x’ = -4y.\,$ Thus if $\,x\,$ is a root then so too is $\,x’ = -4y-x.$

This yields a reflection symmetry $\ \, (x,y) \mapsto (-4y-x,\,y)\,$ on the solution space. Composing

this with the reflection symmetry $\,(x,y)\mapsto (y,x)\,$ yields the map $\,(x,y)\mapsto (-4x-y,x)$

which, iterated starting at solution $(1,0),\,$ yields infinitely many solutions

$$ (1,0),\ (-4,1),\ (15,-4),\ (-56,15),\ (209,-56),\ (-780,209),\ (2911,-78),\ \ldots$$

Sequence $\, 0,1,4,15,56,209,\ldots$ satsifies the recursion $\,f_{n+2} = 4 f_{n+1} – f_{n}\,$ as is easily derived.

This can be transformed to the Pell equation $\ X^2\! – 3 Y^2 = 1\ $ and studied using standard results on Pell equations. See also the comments on this sequence at OEIS sequence A001353.

Hint:

Have you learnt about Pell’s equation?

Try adding a multiple of $y^2$ (on both sides) to complete the square on the left.

Following Aryabhata, write it as $x^2+4xy+4y^2-3y^2=1=(x+2y)^2-3y^2$ Let us define $z=x+2y$, so this becomes $z^2-3y^2=1$. Clearly $z=1,y=0$ is a solution. Now if we have a solution $(z,y)$ we observe that $(z’,y’)=(2z+3y,z+2y)$ is also a solution, because $$\begin {align}z’^2-3y’^2&=(2z+3y)^2-3(z+2y)^2\\&=4z^2+12zy+9y^2-3z^2-12yz-12y^2\\&=z^2-3y^2\\&=1 \end{align}$$ Given any solution we can find a larger one, so there are infinitely many.

A useful resource is this page

It’s easy…
Keep ‘Y’ constant and vary ‘X’.
For single value of ‘Y’ You will get two values of ‘X’,
Go on varying ‘Y’ you will get different value of ‘X’ for each ‘Y’.