No natural numbers sucht that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$. Prove that there are not exist the natural numbers, $x,y,z$ such that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$. 
tell me please if my proof is ok ? 
$x^2+y^2=3\cdot671-3z^2=3(671-z^2)=3k$ but if we want we can find out  $z$. $671 \geq z^2$ so $z=\overline{0\ldots25}.$ 
But we know that a perfect square has the following form: $4k$ or $4k+1$. For example $x^2=4p+1$ and $y^2=4q$ so $x^2+y^2=4(p+q)+1\neq3k$ for $k\neq 3$ and  $p+q \neq 2$. 
I cannot find anothers numbers such that $4(p+q)+1=3k$
Is OK? 
thanks :) 

Just work $\bmod\ 2$: Your first equation reads $x+y+z=0$ and your second reads $x^2+y^2+z^2=1$, but since $x^2\equiv x\bmod\ 2$, the second is equivalent to $x+y+z=1$ and the two are immediately incompatible.
