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This might be an easy question but here goes. I am looking for a polynomial $P\in \mathbb{Q}[x,y,z]$ such that

  1. $P$ is symmetric and homogenous.

  2. $P$ is even in all three variables, i.e. $P\in \mathbb{Q}[x^2,y^2,z^2]$.

  3. $P$ is divisible by $x+y+z$.

In two variables the equivalent of these conditions would be met by $(x^2-y^2)^2$, but I am having trouble constructing one in three variables. Is there some reason why such a polynomial might not exist in three variables?

user82420
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1 Answers1

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Here's one: $$2x^2y^2 + 2y^2z^2 + 2x^2z^2 - x^4 - y^4 - z^4$$ $$= (x + y + z)(x + y - z)(x + z - y)(y + z - x)$$

Moreover, all such polynomials must be divisible by the polynomial above.

To find this, simply note that if $x + y + z$ is a factor, then plugging in $z = -x - y$ results in $0$. But since the polynomial is in $z^2$, plugging in $z = x + y$ gives zero as well. Thus $(x + y - z)$ is a factor too. By symmetry, all the other factors are necessary as well.

You may also recognize this as the square of Heron's formula for the area of a triangle with side lengths $x$, $y$, and $z$ (multiplied by a factor of 16).

Caleb Stanford
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    An algebraic way of discovery is applying the negation maps in each argument to the divisibility relation $(x+y+z)\mid f(x,y,z)$. – anon Jun 14 '13 at 10:21
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    it appears this line of thinking gives you a polynomial satisfying the conditions in any number of variables as well: take the product over all $\pm 1$ sequences of the sum of the variables, where the coefficients match up with the sequence. it's homogeneous of degree $2^n$ – citedcorpse Jun 14 '13 at 10:38