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?

Suppose there is an integer solution. since $y>0$, we have $x-2y < x+2y$.
Hence we have $x-2y =1$
and $x+2y=7$.
if we subtract them, we have $4y = 6$ and $y = \frac32$ which is a contradiction.
Alternatively, think of what values can $x^2 \pmod{4}$ take. Just take modulo $4$ and you can see the contradiction.
