Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.

My Attempt:

We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have

$$ \begin{cases} x^2+y^2\geq 2xy\\ y^2+z^2\geq 2yz\\ z^2+x^2\geq 2zx\\ \end{cases} $$

So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?


Suppose that $(x,y,z)$ is a solution. An even number of these must be odd. If two are odd, say $x$ and $y$, then $x^2+y^2$ has shape $4k+2$, and therefore so does $x^2+y^2+z^2$, since $z^2$ is divisible by $4$. But $2xyz$ has shape $4k$.

So $x,y,z$ are all even, say $2u,2v,2w$. Substituting we get $u^2+v^2+w^2=4uvw$.

Again, $u,v,w$ must be all even.

Continue, forever. We conclude that $x$, $y$, and $z$ are divisible by every power of $2$.

It follows that $x=y=z=0$.

Remark: The same argument can be used for $x^2+y^2+z^2=2axyz$.

This is an instance of Fermat's Method of Infinite Descent, aka induction.


Another method:

It's clear zero is a solution and moreover $xyz\ge0$. Let's prove there are no more solutions over the integers. By the generalized mean theorem:

$$\sqrt{\frac{x^2+y^2+z^2}{3}}>\sqrt[3]{xyz}$$ Which leads to: $$x^2+y^2+z^2 > 3(xyz)^{2/3}$$ So now we know that whenever $3(xyz)^{2/3} > 2xyz$ there are no solutions. Simple algebra shows that this is true whenever $xyz< 27/8$, so the only possible integer solutions are: $$(0,0,0), (0,a,b), (1, 1, 2), (1, 1, 3)$$ And their permutations / sign changes. Now verify these do not satisfy the equation.


  • $\begingroup$ You missed $(0,a,b)$ and permutations. It is easy to show that $a=b=0$ in this case... $\endgroup$ – N. S. Nov 16 '13 at 4:24
  • $\begingroup$ @N.S. valid point :) $\endgroup$ – nbubis Nov 16 '13 at 4:27

There are no solutions. The original equation, considered by Markov, was $$ x^2 + y^2 + z^2 = 3xyz. $$ This leads to the Markov Numbers.

Adolf Hurwitz considered such equations in three or more variables, in 1907. The same reduction procedure as Markov takes any solution to a smaller solution, until one reaches a fundamental solution (grundlosung). There is no fundamental solution with $x^2 + y^2 + z^2 = 2 xyz,$ and no solutions.

In case of interest, see https://mathoverflow.net/questions/84927/conjecture-on-markov-hurwitz-diophantine-equation

Let's see, there is a tree with $x^2 + y^2 + z^2 = xyz,$ but the solutions are not primitive, the fundamental solution is $(3,3,3)$ and so all $(x,y,z)$ are always divisible by $3.$

There is more diversity as the number of variables increases. the first time we get more than one fundamental solution for one of these equations is in $14$ variables, with $$ x_1^2 + x_2^2 + x_3^2 + \cdots + x_{12}^2 + x_{13}^2 + x_{14}^2 = x_1 x_2 x_3 \cdots x_{12} x_{13} x_{14}, $$ which has two fundamental solutions and therefore two trees of solutions, $$ (3,3,2,2,1,1,1,1,1,1,1,1,1,1), $$ $$ (6,4,3,1,1,1,1,1,1,1,1,1,1,1). $$ A collection of trees is (really) referred to as a forest.

Movement within a tree is accomplished by something that is usually called Vieta Jumping on this site.

See a table of fundamental solutions up to $14$ variables at https://mathoverflow.net/questions/142301/a-problem-on-a-specific-integer-partition/142514#142514 where he put the numbers in increasing order instead of decreasing.


We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation.

Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd then, $$x^2+y^2+z^2\equiv(0+0+1)\mod4 ; 2xyz \equiv 0 \mod4$$ So, the only possibility is that all are even.

Let x = 2X , y = 2Y ,z = 2Z. Then, $$4X^2 +4Y^2 +4Z^2 = 16XYZ $$ $$\implies X^2 + Y^2 + Z^2 = 4XYZ$$ The same argument goes show that X,Y,Z are even. The process can be continued indefinitely.

This is possible only when x = y = z = 0. So ,there exist no other solution other than this.


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