Proving $x^2 - 4y^2 = 7$ has no natural numbers

Question

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?

Answer 1

Hint As you said $7=7 \cdot 1$ or $7=(-7)\cdot (-1)$. Therefore, the only possibilities are $$x-2y=1 \\ x+2y=7$$

or $$x-2y=-7\\ x+2y =-1$$

Solve those systems and see what happens.

Answer 2

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.

Answer 3

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.