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

Solution 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 $0$.

Solution 2:

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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Solution 3:

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} $