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

Question

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??

Answer 1

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 $0$.

Answer 2

$2xyz$ 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.

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

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

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

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 $\Bbb{N}$

Answer 3

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.

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