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.
 A: 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 $1$.
A: For the congruence to hold, $y$ and $z$ must have fifferent parities.
If $y$ is odd and $z$ is even, then $y^2\equiv 1\pmod{4}$, and therefore $3y^2\equiv 3\pmod{4}$, and therefore $3y^2-z^2\equiv 3\pmod{4}$.
If $y$ is even and $z$ is odd, then $z^2\equiv 1\pmod{4}$, and therefore $3y^2-z^2\equiv -1\equiv 3\pmod{4}$. 
Remark: We can symmetrize, by noting that $3y^2\equiv -y^2\pmod{4}$, and therefore $3y^2-z^2\equiv -(y^2+z^2)\pmod{4}$. The advantage is that $y^2+z^2$ is symmetric in $y$ and $z$, so without loss of generality there is only one case to consider. 
A: mod $\,4\!:\,\ \color{#c00}{{-}1} \not\equiv -x \equiv  y^2\!+x^2 \equiv \{0,1\}\!+\!\{0,1\} \equiv \{0,1,2\}\ \ $ QED
