Prove the diophantine equation $x^2-7y^2=67$ has no solutions I've attempted using modulo to try and solve this, i.e. by doing the following:
$$x^2-7y^2=67 \implies x^2=4 \in \mathbb{Z}_7$$
That would imply there are solutions, but that isn't the case so I'm at an impasse.
Any help is appreciated.
 A: $$x^2 - 7 y^2 \equiv x^2 + y^2 \pmod 4.$$
That is, $x^2 - 7y^2 \neq 3 \pmod 4$
A: Another way to look at this is to ask if $(x - y \sqrt 7)(x + y \sqrt 7) = \pm 67$ has solutions in integers.
This gets into algebraic number theory, which is a subject you might really enjoy or really be frustrated by.
If the equation has no solutions, we say that $67$ is prime in $\mathbb Z[\sqrt 7]$. The Tooth Fairy and Legendre tell us that $$\left(\frac{7}{67}\right) = 7^{33} \equiv -1 \pmod{67},$$ which means the equation $x^2 - 7y = 67$ has no solutions and neither does $x^2 - 7y^2 = 67$. As $\mathbb Z[\sqrt 7]$ is a unique factorization domain, $67$ is indeed prime in this domain.
For the sake of comparison, solve $x^2 - 7y^2 = -83$ in integers. You will find that $(13 - 6 \sqrt 7)(13 + 6 \sqrt 7) = -83$. As the fundamental unit of this domain has norm $1$ rather than $-1$, you will also find $x^2 - 7y^2 = 83$ has no solutions, a wrinkle that will either delight you or infuriate you.
A: Note that one of them is odd and other must be even, an odd perfect square leaves a remainder $1$ when divided by $4$, so if $x$ is odd $x^2-7y^2$ leaves a remainder $1$ when divided by $4$, and if $y$ is odd also leaves a remainder $1$ when divided by $4$, hence it can never equal $67$ as it leaves a remainder $3$ when divided by $4$
