There is no square number that is $3 \mod 4$

Question

Prove that the following cannot be true: There is no square number that is $3 \mod 4$

$x^2 \equiv 3 \mod 4$, I started with examples:

$1^2 \mod 4=1$

$2^2 \mod 4=0$

$3^2 \mod 4=1$

I am more interested in the proof but Im not seeing how to do it.

Answer 1

If $x$ is even then $x=2k$ so the square of $x$ is equal $4k^2$ and leaves no remainder when divided by $4$.

If $x$ is odd then $x=2k-1$ so the square of $x$ is equal $4k^2-4k+1$ and leaves a remainder $1$ when divided by $4$.

So we have either $x^2 \equiv 0 \mod 4$ or $x^2 \equiv 1 \mod 4$.

Answer 2

$0^2=0$, $1^2=1$, $2^2=4\equiv 0$, $3^2=9\equiv1$. This looks like an example, but it is in fact a proof...

Answer 3

First of all make the obvious observations that if $x^2 \equiv 3 \bmod 4$ then $x^2$ is odd, and $x$ must be odd as well.

Then there are two possibilities modulo $4$: either $x \equiv 1 \bmod 4$ or $x \equiv 3 \bmod 4$. If the former, then $x^2 \equiv 1 \bmod 4$ (no surprise there, right?). If the latter, then $x^2 \equiv 9 \bmod 4$ (since $3 \times 3 = 9$), but we can reduce that further to $x^2 \equiv 1 \bmod 4$.

Clearly even $x$ won't work, so this takes care of all integer possibilities for $x$.