modular arithmetic proof Suppose $x$, $y$, and $z$ are integers and $x= 3y^2 -z^2$.  Prove that $x\not\equiv1\mod4$.
My thoughts: So I am not sure the route that can prove this. I am trying to just use the simple stuff to set me up.  I started with saying if $x\equiv1\mod4$ you would have $4|x-1$ and there will exist an integer, say $t$, where $x-1 = 4t$.  I then solved for $x$ just because the other equation was in terms of $x$.  So from that we would have $x=4t+1$.  Not sure if I am on the right track here or what I can do with what I have.

The base is small enough for this problem that we can use proof by exhaustion by letting $y,z \equiv 0,1,2,3 \pmod 4$. And in fact, because $0^2 \equiv 2^2 \pmod 4$ and $1^2 \equiv 3^2 \pmod 4$, you only have to check $4$ different cases:


*

*$y \equiv 0, z \equiv 0 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 0^2 - 0^2 \equiv 0$

*$y \equiv 0, z \equiv 1 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 0^2 - 1^2 \equiv 3$

*$y \equiv 1, z \equiv 0 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 1^2 - 0^2 \equiv 3$

*$y \equiv 1, z \equiv 1 \pmod 4 \Rightarrow 3y^2 - z^2 \equiv 3 \cdot 1^2 - 1^2 \equiv 2$
And none of these are $1$.
