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Does $$x^2+y^2=3(z^2+ u^2)$$ have solutions in positive integers? I was assigned this problem, but I am struggling to find a solution. I guess that a proof by contradiction is required.

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up vote 2 down vote accepted

Suppose, for the sake of contradiction, that the equation $$x^2+y^2=3(z^2+u^2)$$ has solutions in positive integers. Then among them must be one, which we will denote by $(x_1,y_1, z_1, u_1)$, with the smallest possible value of $x$. From the equation $$x_1^2+y_1^2=3(z_1^2+u_1^2)$$ it follows that $x_1^2+y_1^2$ is divisible by $3$. Hence, both $x_1$ and $y_1$ are divisible by $3$ (indeed, if an integer is not divisible by 3, then its square is of the form $3k+1$ for some integer $k$). Therefore,$x_1=3x_2$ and $y_1=3y_2$ for some positive integers $x_2$ and $y_2$. It follows that $$9x_2^2+9y_2^2=3(z_1^2+u_1^2),$$ that is $$3x_2^2+3y_2^2=z_1^2+u_1^2.$$ This implies that $u_1$ and $z_1$ are divisible by $3$. Therefore, $z_1=3z_2$ and $u_1=3u_2$ for some positive integers $z_2$ and $u_2$. It follows that $$x_2^2+y_2^2=3(z_2^2+u_2^2).$$ This means that $(x_2,y_2, z_2, u_2)$ is a solution of the first equation. However, $x_2= \frac{1}{3}x_1 < x_1$, which contradicts the assumption that $x_1$ was the smallest possible value for all solutions of the equations. Hence, the equation has no solution in positive integers.


This problem is discussed in J. Cofman, What to Solve? Problems and Suggestions for Young Mathematicians, Oxford University Press, 1990.

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Nice one. Thank you. Also, I'll check that book :). – user140619 Apr 5 '14 at 20:17
You're welcome. – user139587 Apr 5 '14 at 20:26


$$x^2+y^2=0\pmod 3\iff\;\;\begin{cases}x=0\pmod 3\\y=0\pmod 3\end{cases}$$

But then

$$9\mid(x^2+y^2)\;\implies 3\mid(z^2+u^2)$$

Well, check those powers of three in both sides and get a contradiction...

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Thank you for your suggestions. I really appreciate them. – user140619 Apr 5 '14 at 20:16

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