I have following equation: $$x^2 + y^2 + z^2 + v^2 = 2 x y z v.$$ How can I solve this one? Roots should be integer numbers.

Edit: I tried to prove, that given equation don't have solutions. I fixed $z, v$ variables. Then task boils down to finding intersection between paraboloid and saddle. Unfortunately, it is hard way. Maybe there are another ways/patterns for solving similar equations?

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    $\begingroup$ Александр, здесь правилами сайта оговорено, что Вы должны показать свои попытки решения этой задачи, иначе вопрос будут минусовать и закрывать $\endgroup$
    – Roman83
    Commented Aug 3, 2016 at 16:58

2 Answers 2


Outline: It will turn out that the only solution is with all the variables equal to $0$.

The right-hand side is even. So it is clear our variables are not all odd.

If two are odd and two are even, the left-hand side is congruent to $2$ modulo $4$, but the right-hand side is divisible by $4$, impossible.

So all the variables are even. Let $x=2x_1$, $y=2y_1$, and so on.

Substitute in our equation and simplify. We get $$x_1^2+y_1^2+z_1^2+v_1^2=8x_1y_1z_1v_1.$$

Now repeat the argument. We find that $x_1,y_1,z_1,v_1$ are all even. Let $x_1=2x_2$, $y_1=2y_2$, and so on. We get $$x_2^2+y_2^2+z_2^2+v_2^2=32x_2y_2z_2v_2.$$

Continue. We conclude that each of $x,y,z,v$ is divisible by arbitrarily high powers of $2$, so all of them are $0$.

Remark: This sort of argument is usually called an infinite descent argument. It is ordinarily credited to Fermat. It is essentially an induction argument.

If you prefer, you can write it up as follows. If there is a solution with the variables not all equal to $0$, then there is a solution with $|x|+|y|+|z|+|v|$ positive and minimal. But then $(x_1,y_1,z_1,v_1)$ is a solution with sum of the absolute values greater than $0$ but less than $|x|+|y|+|z|+|v|$: Contradiction!

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    $\begingroup$ Brilliant description $\endgroup$ Commented Aug 3, 2016 at 17:14

On the other hand, there are infinitely many solutions to $$ x^2 + y^2 + z^2 + v^2 = 4 xyzv,$$ occurring in a tree similar to the tree of Markov Numbers. It took me a while to find out, but most Markov things are the same person.

There are also infinitely many solutions to $$ x^2 + y^2 + z^2 + v^2 = 4 xyzv,$$ although this time all the variables are even, dividing all of them by $2$ gives a quadruple in the first tree.

For details, see HURWITZ

  • $\begingroup$ Interesting approach. Unfortunately I cannot understand language of article (I am native Russian) $\endgroup$ Commented Aug 3, 2016 at 20:51
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    $\begingroup$ @АлександрЛысенко then I will just describe it. To solve $$ x_1^2 + \cdots x_n^2 = x \, x_1 x_2 \cdots x_n $$ in positive integers, we must have $x \leq n,$ also every tree of solutions has a "fundamental" solution with $$ \max {2 x_i^2} \leq x \, x_1 x_2 \cdots x_n $$ He gives enough inequalities so that it is a finite search to find fundamental solution for a fixed $a,x$ or show nonexistence. $\endgroup$
    – Will Jagy
    Commented Aug 3, 2016 at 20:58

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