I want to show that the equation
$x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent.

This question has been solved with infinite descent by showing that $x,y,z$ are all even and we can infinitely find smaller solutions: Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$

Can it be shown instead by assuming you have the least positive solution and then finding a smaller solution.
I let ($a,b,c$) be the least positive solution with gcd(a,b,c) = 1, and I want to find a smaller solution.
I have that exactly one of $a,b,c$ must be even because if two were even, then the left hand side would be odd while the right hand side is even. And if all three are even, then gcd($a,b,c$) $\ge$ 2.

So we can assume $a$ is even. I've tried considering the equation mod $a,b,c,2$ but couldn't seem to find a way to construct a smaller solution.

  • $\begingroup$ Um, isn't the accepted solution of the first link you provided using infinit descent? $\endgroup$
    – SK19
    May 13, 2018 at 18:42
  • $\begingroup$ @SK19 You are right, I have edited the question to be clearer about what I am looking for. $\endgroup$
    – mpears
    May 13, 2018 at 18:57
  • $\begingroup$ Try mod 4 for a start. $\endgroup$
    – Old Peter
    May 13, 2018 at 19:04
  • $\begingroup$ original proof in Hurwitz (1907) zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf The technique is usually called Vieta Jumping on this website, but Hurwitz adds some useful ideas. $\endgroup$
    – Will Jagy
    May 13, 2018 at 19:12

2 Answers 2



in this equation $RHS$ is even .then, $LHS$ is also even.

w.l.o.g, $z$ is even $\exists z_1 $ such that $z= 2 z_1$ and $x,y$ are in same parity.

substituting $z= 2 z_1$ in given equation

$x^2+y^2+4z_1^2=4xyz_1 ........................(2)$


$RHS$ is multiple of 4 .then ,$LHS$ must be multiple of 4

therefore $x$ and $y$ are even.

let $x=2x_1$ and $y=2y_1$

substituting $x=2x_1$ and $y=2y_1$ in $(2)$ we get,

$4x_1^2+4y_1^2+4z_1^2=16x_1y_1z_1 $

$\implies x_1^2+y_1^2+z_1^2=4x_1y_1z_1 $

$LHS$ is multiple of $4$.then,$RHS$ is also multile of $4$

w.l.o.g,$z_1$ is even, let $z_1=2z_2$

$\implies x_1^2+y_1^2+4z_2^2=8x_1y_1z_2 $

in similar argument .We can find three sequence of infinite integers

$$x_1 \gt x_2 \gt x_3 \dots \gt x_n$$

$$y_1 \gt y_2 \gt y_3 \dots \gt y_n$$

$$z_1 \gt z_2 \gt z_3 \dots \gt z_n$$

that can solve the equation $x^2+y^2+z^2=2xyz$.

by using infinite descent we can prove $x^2+y^2+z^2=2xyz$ has no solution greater than $0$.


You basically want to turn a proof of infinite descent into a "simple" proof by contradiction.

Infinite descent basically works like this:

Typically, one shows that if a solution to a problem existed, which in some sense was related to one or more natural numbers, it would necessarily imply that a second solution existed, which was related to one or more 'smaller' natural numbers. This in turn would imply a third solution related to smaller natural numbers, implying a fourth solution, therefore a fifth solution, and so on. However, there cannot be an infinity of ever-smaller natural numbers, and therefore by mathematical induction (repeating the same step) the original premise—that any solution exists— is incorrect: its correctness produces a contradiction.

You want something like this:

An alternative way to express this is to assume one or more solutions or examples exists. Then there must be a smallest solution or example—a minimal counterexample. We then prove that if a smallest solution exists, it must imply the existence of a smaller solution (in some sense)—which again proves that the existence of any solution would lead to a contradiction.

(Citations from WP) As it is indicated in the text, this is just a reformulation of said proof. If I were to borrow André's argumentation, I would write:

Suppose that $(x,y,z)$ is a smallest solution (i.e. $\gcd(x,y,z)=1$). An even number of these must be odd (this can be seen by taking the equation mod 2). 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 $4$ (because $z$ is even). But $2xyz$ has shape $4k$. So $x,y,z$ are all even, say $2u,2v,2w$. But this contradicts our example of a smallest solution, and we are done.

This is the step you have probably missed in André's solution:

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 $4k$.

With this you could have finished your proof. Alas, I don't see this as duplicate, but rather as proof-explanation.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .