Prove $4x^2 - y^2 = 1$, for $x,y \in \mathbb N$, has no solutions by contradiction

Question

I'm trying to prove $4x^2 - y^2 = 1$, for $x,y \in \mathbb N$, has no solutions by contradiction. Assume that $4x^2 - y^2 = 1$, for $x,y \in \mathbb N$, has a solution. I've tried to arrive at a contradiction by finding an explicit formula for $x$ in terms of $y$ and trying to use the difference of two squares but none of these methods have worked. I've tried using the fact that if we let $x^2=m$ and $y^2=n$ then we can say $gcd$($m,n$) = $1$ which also means $gcd$($x,y$) = $1$ but I also can't see if this leads me to a contradiction either.

Answer 1

You can write it as $(2x+y)(2x-y)=1$ which can only be true if $2x+y=2x-y=\pm 1$, but this gives $y=0, x=\pm \frac 12$, which is not an integer. Alternately, write it as $(2x)^2-y^2=1$ and note that the only pair of sequential squares is $0,1$ with the same result.

Answer 2

A square is always $\equiv 0$ or $\equiv 1$ (mod $4$)

Since $4x^2 \equiv 0$ (mod $4$), you get $y^2 \equiv 4x^2-1 \equiv 3$ (mod $4$) which is impossible.

Answer 3

$4x^2$ is even. $1$ is odd. So $y^2=4x^2-1$ is odd. So $y $ is odd.

Let $y=2n+1$. So $4x^2-y^2=4x^2-4n^2-4n-1=1$

So $4x^2-4n^2-4n=2$. So $x^2-n^2-n=\frac 12$.

But $x,y $ and $x^2-n^2-n $ are integers. This is a contradiction.

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Which is equivalent to the standard proof, the perfect squares are congruent to $0\mod 4$ if even, and to $1\mod 4$ if odd, but never to $2$ or $3 \mod 4$.

The proof is nearly identical.