Proving $x^2 - 4y^2 = 7$ has no natural numbers Ok so I needed to prove this by contradiction.
Let $P:~x^2 - 4y^2 = 7$ and $Q:~x,y$ are not natural numbers
Note that $N$ does not include $0$ 
OK to begin to prove by contradiction we are given $P$ and I would assume $\text{not}(Q)$ so the equation would have natural numbers
$$x^2 - 4y = 7 = (x-2y)(x+2y) = 7$$ 
Ok so here $(x-2y)(x+2y) = 7$  would be the contradiction but I don't seem to understand why,  I know that $7$ could be written as a product of two natural numbers in 2 ways, so because we only have one way is that why we would say that $(x-2y)(x+2y) = 7$  is a contradiction?

Another approach: 
Just observe that, left hand side will leave $0$ or $1$ as remainder when you will divide the left side by $4$. Because, $4$ divides $4y^2$, and it is well known that when you divide any square by $4$ it leaves $0$ or $1$ as a remainder. So, $x^2-4y^2$ will be either of $4k$ or $4k'+1$ form. But, $7=4+3$, so rhs will leave $3$ as a remainder. Hence, we can't have any integer which satisfies this equation.
