Prove that there exists no natural number $x$ such that $x^2-6$ is a perfect square

Question

I tried to prove this question using contradiction. I first assumed that there is such a perfect square and then claimed that any perfect square can be expressed in $n^2$, where $n$ is an integer, then I said that any perfect square is expressed as $(n-a)^2$, and factored $a^2-6$ using the quadratic formula and showed that $a$ can not be a natural number. This is a valid proof right?

Answer 1

Hint

$$x^2-6=y^2 \Rightarrow x^2-y^2=6$$

Now use the difference of squares formula, and solve.

Note that if you look at the parities of $x-y, x+y$ you don't need to solve.

Answer 2

A perfect square is always either $0 \bmod 4$ or $1 \bmod 4$.

Therefore either $x^2 - 6 \equiv 0-6 \equiv 2 \pmod 4$ or $x^2-6 \equiv 1-6 \equiv 3 \bmod 4$, neither of which is possible for a perfect square.