# Show that there are no nonzero solutions to the equation $x^2=3y^2+3z^2$

I am asked to show that there are no non zero integer solutions to the following equation $x^2=3y^2+3z^2$

I think that maybe infinite descents is the key.

So I started taking the right hand side modulo 3 which gives me zero. Meaning that $X^2$ must be o modulo 3 as well and so I can write $X^2=3k$ , for some integer K and (k,3)=1.

I then divided by 3 and I am now left with $k=y^2+z^2$ . Now I know that any integer can be written as sum of 2 squares if and only if each prime of it's prime factorization has an even power if it is of the form 4k+3. But yet I am stuck . If anyone can help would be appreciated.

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You can write $x=3k$ so $x^2=9k^2$ then divide by $3$. Then argue both $y,z$ are multiples of $3$.

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Assume $\,x,y,z\,$ have no common factor. Now let us work modulo $\,4\,$ : every square is either $\,0\,$ or $\,1\,$ , and since

$$3y^2+3z^2=-(y^2+z^2)=\begin{cases}0\,\;\;,\,y,z=0,2\\{}\\-1=3\,\;\;,\,y=1\,,\,z=0\,\,or\,\,y=0\,,\,z=1\\{}\\-2=2\;\;\,,\,y=z=1\end{cases}$$

so the only possibility is the first one, and thus also $\,x^2=0\,$ , but then all of $\,x,y,z\,$ are even...

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The exponent of 3 in $y^2 + z^2$ is automatically even. If $y^2 + z^2 \equiv 0 \pmod 3,$ since the possible values of $y^2$ or $z^2$ modulo 3 are just 0,1, it follows that both must be 0. That is, $y^2 + z^2 \equiv 0 \pmod 3 \Longrightarrow y,z \equiv 0 \pmod 3.$ So, actually, $y^2 + z^2$ is divisible by 9. Now, take $y_1 = y/3, \; z_1 = z/3.$ If $y_1^2 + z_1^2$ is not divisible by 3, that was it, the original was divisible by $9 = 3^2.$ Otherwise, $y_1^2 + z_1^2 \equiv 0 \pmod 3 \Longrightarrow y_1,z_1 \equiv 0 \pmod 3,$ take $y_2 = y_1/3, \; z_2 = z_1/3.$ Continue until some $y_j^2 + z_j^2$ is not divisible by 3. Then $3^{2j} | (y^2 + z^2),$ or more precisely $$\mbox{ord}_3 (y^2 + z^2) = 2 j.$$

It seems the OP wrote in part of this. Next we get $$\mbox{ord}_3 \; 3(y^2 + z^2) = 2 j + 1.$$ But $$\mbox{ord}_3 \; x^2 = 2 k$$ for some $k.$

EEDDDIITTTT: I think I get it. The OP mentioned descent, in which case it is enough to show that $x,y,z$ must all be divisible by 3, as you can then divide through by 9 and keep going. A lifestyle choice. I try not to judge.

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If $3|x^2$ then, $9|x^2$ too $\implies 3|(y^2+z^2)$ and you can make it from here.

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How about looking at solutions where $y=z$, then $x^2=6y^2$ and so $y=z=\frac{1}{\sqrt{6}}x$. Now let $y=z$ be an arbitrary number $a$ thus implying $x=\sqrt{6}a$ and there are infinitely many solutions $(x,y,z)=(\sqrt{6}a,a,a)$.