Triple Trouble

Number Theory Level 3

$x^2 + y^2 + z^2 = 2xyz$

How many unordered triplets $(x,y,z)$ (such that $x$ , $y$ and $z$ are positive integers) exist such that the equation above is satisfied.

The answer is 0.

1 solution

Brian Charlesworth
Jul 14, 2015

We must either have exactly two of $x,y,z$ odd or all of them even. If exactly two are odd, say $x,y,$ then $x^{2} + y^{2} \equiv 2 \pmod{4},$ and since $z$ is even we would also have $x^{2} + y^{2} + z^{2} \equiv 2 \pmod{4}.$ But in this scenario $2xyz \equiv 0 \pmod{4},$ and thus any solution must have $x,y,z$ all even.

So let $x = 2a, y = 2b, z = 2c.$ Then the equation becomes

$4a^{2} + 4b^{2} + 4c^{2} = 16abc \Longrightarrow a^{2} + b^{2} + c^{2} = 4abc.$

Similar to before, either exactly two of $a,b,c$ are odd or all of them are even. We can again conclude that all are even, say $a = 2k, b = 2m, c = 2n.$ We then end up with the equation

$4k^{2} + 4m^{2} + 4n^{2} = 32kmn \Longrightarrow k^{2} + m^{2} + n^{2} = 8kmn.$

We can then continue with this process ad infinitum, (i.e., infinite descent ), resulting in the observation that $x,y,z$ are each divisible by every power of $2.$ Since $x,y,z$ must be finite, the only possibility left is that $x = y = z = 0,$ implying that there are $\boxed{0}$ solution triples of positive integers.

I'd just like to add equations of the form $x^2+y^2+z^2=nxyz$ are called markov-hurwitz equations.

I find markov numbers to be a fun read.

Isaac Buckley - 5 years, 11 months ago

Huh, I didn't know about these. I wouldn't have guessed that for $n = 3$ there was an infinite solution set. For $n = 1$ we have $(x,y,z) = (3,3,3), (6,3,3), (3,6,3), (3,3,6),$ and no solutions $n \ne 1,3.$ More generally we would be looking at the equation

$\displaystyle\sum_{i=1}^{n} x_{i}^{2} = a\prod_{i=1}^{n} x_{i}.$

Apparently there are no solutions when $a \gt n.$ For $a = n$ there is always the solution $(1,1,1,.....,1,1),$ and (I think) an infinite set of other solutions.

Variations of this equation that might be interesting to consider are

$\displaystyle\sum_{i=1}^{n} x_{i}^{3} = a\prod_{i=1}^{n} x_{i}$ and $\displaystyle\left(\sum_{i=1}^{n} x^{i}\right)^{2} = a\prod_{i=1}^{n} x_{i}.$

Or maybe not .... :)

Brian Charlesworth - 5 years, 11 months ago

Well for $n=1$ we have $x^2+y^2+z^2=xyz$ .

Making the sub $x=3a,\,y=3b,\,z=3c$

It turns into $a^2+b^2+c^2=3abc$ . So the solutions to $n=1$ are just 3 times the markov numbers.

I think you might be able to get an infinite amount if you can find $1$ solution.

You can show for $n=3$ That if $(x,y,z)$ is a solution then so is $(x,y,3xy-z)$ to generate an infinite amount in this case and there might be something similar you can do in other cases to generate solutions from one solution.

I might just have to play with those new equations now :P

Isaac Buckley - 5 years, 11 months ago

@Isaac Buckley – Yes, I should have noticed that substitution. With the Markov number tree it does seem that if you can find one solution then the solution set is infinite. There are some interesting connections between Markov numbers and Fibonacci numbers, as well as Pell numbers. Cool. Thanks for the info. :)

Brian Charlesworth - 5 years, 11 months ago

Triple Trouble

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