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I'm working on a problem where we must show that there are infinitely many integer triples $x,y,z$ such that $x^2 + y^2 + z^2$ is divisible by $(x + y +z)$ and $x,y,z$ are pairwise coprime. Also $x,y,z$ are distinct.

I can see that either all $x,y,z$ are odd or that one of them is even but besides that haven't made much progress (I can't see any nice factorization of $x^2 + y^2 + z^2$ nor have I been able to find an easy example to get a foothold).

Broadly I'm thinking that my two major approaches are proof by contradiction that the number of triples can't be finite, or trying to find some construction to generate the triples.

I imagine finding a construction would be more fruitful.

I would appreciate both a solution but also advice on how to approach this type of problem

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  • $\begingroup$ I'm wondering if the pigeonhole principle is involved. $\endgroup$ – Michael Hardy Jan 15 '18 at 7:57
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Take $(x, y, z) = (3k, 3k + 1, 54k^2 + 12k + 1)$ for arbitrary $k \in \mathbb{N}_+$. Since$$ (x, y) = (3k, 3k + 1) = 1, $$\begin{align*} (x, z) &= (3k, 54k^2 + 12k + 1)\\ &= (3k, 54k^2 + 12k + 1 - 3k(18k + 4)) = (3k, 1) = 1, \end{align*}\begin{align*} (y, z) &= (3k + 1, 54k^2 + 12k + 1)\\ &= (3k + 1, 54k^2 + 12k + 1 - (3k + 1)(18k - 2)) = (3k + 1, 3) = 1, \end{align*} then $x, y, z$ are pairwise coprime. Note that $x + y + z = 54k^2 + 18k + 2$ and\begin{align*} x^2 + y^2 + z^2 &= (3k)^2 + (3k + 1)^2 + (54k^2 + 12k + 1)^2\\ &= 2916k^4 + 1296k^3 + 270k^2 + 30k + 2\\ &= (54k^2 + 18k + 2)(54k^2 + 6k + 1), \end{align*} thus $x+ y +z \mid x^2 + y^2 + z^2$.

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This is more than a comment, but not a whole solution:

You can analyse this as follows:

$$x^2+y^2+z^2=a(x+y+z)$$

Multiply by $4$ and complete the square: $$(2x-a)^2+(2y-a)^2+(2z-a)^2=3a^2$$

For each $a$ there can be only a finite number of solutions, one of which is the inadmissible $x=y=z=a$. So you are looking for numbers of the form $3a^2$ which can be expressed as the sum of three squares in more than one way, and you have a parity and divisibility considerations too.

With $a=5$ you get $x=3, y=5, z=6$ which doesn't fit your pairwise coprime condition. But this gives a start on finding all solutions.

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