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

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.

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.

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