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!
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.
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'.