Prove that $n^2-4m-2\ne0$ by contradiction

Question

Let $n$ and $m$ be integers. How do I prove that $n^2-4m-2$ is never zero, by contradiction? I have no idea how to start.

Answer 1

HINT :

Show that no square is of the form $4m+2$ By considering that it is divisible by 2 and not by 4.

Answer 2

An elementary application of congruences follows. Suppose on the contrary that a solution exists, i.e. $$n^2=4m+2$$ for some $m,n\in\Bbb Z$. Then, considering the equation modulo 4:

• the left-hand side is either 0 or 1 modulo 4
• the right-hand side is 2 modulo 4

But $0\ne2$ and $1\ne2$, so we always get a contradiction. Hence the equation has no solutions in integers.

Answer 3

Suppose $n^2 = 4m+2$. $4m+2$ is even, so $n^2$ is even, so $n$ is even, say $n=2k$. Then $n^2 = 4k^2 = 4m+2$. Then $2 = 4m-4k^2 = 4(m-k^2)$. So $2$ is a multiple of $4$. Contradiction.