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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 :)

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$3|(4(p+q)+1)$ is possible... – ougao Dec 14 '12 at 22:54
I think you should instead use that $x^2$ is congrunt to 0, 1 mod 3 – ougao Dec 14 '12 at 22:56
Where is this problem from? I get nervous when I see calendar years involved. – Will Jagy Dec 14 '12 at 22:58
For your last, if $p=q=1$, then $4(p+q)+1=9=3\cdot 3$ – Ross Millikan Dec 14 '12 at 23:00
@WillJagy it is a problem for the most important mathematical magazine from Romania(gazeta matematica) – Iuli Dec 14 '12 at 23:23
up vote 4 down vote accepted

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.

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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 $12$.

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I think that the notation means that $z$ is an integer between $0$ and $25$, since $26^2=676$, but $671-z^2>0$. – Asaf Karagila Dec 14 '12 at 23:03

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