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 :)