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Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$

$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??

Hamid Reza Ebrahimi
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4 Answers4

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We use a classical infinite descent argument.

Note that the right-hand side is even, so the left-hand side must be. It follows that two of $x$, $y$, $z$ are odd and the third even, or all three are even.

But two odd and one even is impossible, for then the right-hand side is divisible by $4$ and the left-hand side is not.

Thus $x=2x_1$, $y=2y_1$, $z=2z_1$ for some integers $x_1$, $y_1$, $z_1$.

Substituting we get $x_1^2+y_1^2+z_1^2=4x_1y_1z_1$.

Repeat the argument. We find that $x_1=2x_2$, and so on, with $x_2^2+y_2^2+z_2^2=8x_2y_2z_2$.

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

André Nicolas
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  • Is there any other heuristic method for this question? – Hamid Reza Ebrahimi Jul 07 '16 at 05:24
  • @HamidRezaEbrahimi: I do not know of a substantially different approach. We can rephrase the argument in standard descent terms, $x_1,x_2,\dots$ an infinite descending chain of positive integers, which is impossible. Or we can rephrase it as an induction argument. – André Nicolas Jul 07 '16 at 05:31
  • @AndréNicolas Sorry, I made quite a big blunder there! – Jeevan Devaranjan Jul 07 '16 at 05:46
  • Andre, in case of interest, A. Hurwitz considered exactly this in 1907, that is $$ x_1^1 + x_2^2 + \cdots x_n^2 = a x_1 x_2 \cdots x_n, $$ mostly for $n \geq 3.$ Maybe I should post his table for $n \leq 10.$ For a fixed $n,a$ all solutions make up a forest, movement between solutions in a tree is by "Vieta Jumping." – Will Jagy Jul 07 '16 at 18:18
  • @WillJagy: Thank you, I knew there was a literature. But though I can kind of remember what I have seen, where I have seen it has never been easy, and is not getting easier. Of course the particular case of this problem is not an issue, we have all assigned variants of it in elementary number theory courses. – André Nicolas Jul 07 '16 at 18:38
  • Andre, my websites have been revived, in shortened forms. Here is the Hurwitz article http://zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf It really is worthwhile, the assistance for most contest type problems is to say "Vieta Jumping as contest people call it, is just the first step, after which we have a set of inequalities defining a region to search for integer solutions..." This has been very helpful to me. – Will Jagy Jul 07 '16 at 18:45
  • Using the Hurwitz viewpoint, very proud of this one, http://math.stackexchange.com/questions/829228/is-it-true-that-fx-y-dfracx2y2xy-t-has-only-finitely-many-distinct-i although it took a few tries and I still needed some help – Will Jagy Jul 07 '16 at 18:47
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    @WillJagy: Thank you, bookmarked. – André Nicolas Jul 07 '16 at 18:52
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In 1907, A. Hurwitz considered $$ x_1^2 + x_2^2 + \cdots + x_n^2 = x \, x_1 x_2 \cdots x_n $$ The conclusions, with all the $x_i \geq 0$ integers and $x$ an integer, included $x \leq n.$ The main thing, though is that, for a fixed pair $n,x,$ all solutions collected into a finite set of rooted trees. Travel within a tree is by "Vieta Jumping." He called a tree root a Grundlösung, I usually say fundamental solution. He gave enough inequalities to find all fundamental solutions for any given pair $n,x,$ including showing when there were no actual solutions using that pair.

making jpeg, just a minute

enter image description here

Will Jagy
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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 over $\mathbb{N} $

Identicon
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$2xyz$ even $\rightarrow$ $x$ even and $y$,$z$$\in$ odd OR $x$,$y$,$z$$\in$ even

  1. $x=2k$ $y$,$z$$\in$odd:

$4k^2 + y^2 + z^2 = 4kyz$

$4k(k-yz) +y^2 + z^2 = 0$

We have this problem: $4|y^2 + z^2$ so $y$ or $z$ must be even but are odd.

  1. $x=2k, y=2l, z=2m$:

$4k^2 + 4l^2 + 4m^2 = 16klm$

$k^2 + l^2 + m^2 = 4klm$

And the problem is infinite (problem is the same like at the beginning - 4 instead of 2 nothing change) but $x,y,z$ should be finite.

There is no answer over $\Bbb{N}$

Piotr Wasilewicz
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