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 :) 

You are fine to $x^2+y^2=3(671-z^2)=3k$, but I don't know what you mean by $z=\overline{0\ldots25}$.  You are correct that perfect squares are $4k$ or $4k+1$ (but be careful about reusing $k$.  You might confuse yourself thinking they are the same).  This means $x^2+y^2 \equiv 0,1,2 \pmod 4$ but that doesn't say anything about modulo $3$.  In fact, $x^2+y^2$ can be anything $\pmod 3$  If you want to combine modulos $3$ and $4$ you can work modulo $12$.
