Taking a power of a polynomial to make it symmetric Suppose I have a non-symmetric multi-variable polynomial in $n$ variables $P(x_1,x_2,...,x_n)$. For example $P$ might be $x_1^2+x_2$ or $x_1-x_2$
Under what conditions will some power $m$ of $P$ (that is $P(x_1,x_2,...,x_n)^m$) be a symmetric polynomial in $x_1,x_2,...,x_n$?
For example it seems like there is no such $m$ in the case of $x_1^2+x_2$.
But there is such an $m$ in the case of  $x_1-x_2$ since $(x_1-x_2)^2=x_1^2-2x_1x_2+x_2^2$.  
 A: First of all: this will happen if and only if every permutation of the variables multiplies $P$ by some root of unity.
If $P\circ\sigma = \zeta_\sigma P$ for every permutation $\sigma$, then $P^n$ will be symmetric, where $n$ is the least common multiple of the orders of the $\zeta_\sigma$, because $P^n \circ \sigma = (P\circ \sigma)^n = (\zeta_\sigma P)^n = \zeta_\sigma^n P^n = P^n$.
On the other hand, if $P^n$ is symmetric, then, for any permutation $\sigma$, we have $(P\circ \sigma)^n = P^n$, so $P\circ \sigma$ and $P$ have the same irreducible factors with the same multiplicity, hence $P\circ \sigma = aP$ for some constant $a$ with $a^n = 1$.
But this cannot really happen if the full symmetry group is acting!—since a symmetry group is generated by transpositions, we can have at most order $2$ behavior.  And since two transpositions generate a symmetry of order $3$, there are only two possibilities: either $P$ is symmetric already, or it is antisymmetric in every pair of variables (e.g. $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$).
These examples are fairly interesting, actually—they are Vandermonde determinants, which pick out the sign of the acting permutation.  Specifically, the antisymmetric polynomials are those which are the product $\prod_{i<j} (x_i -x_j)$ times any symmetric polynomial.
If you require only cyclic symmetry, then you will get interesting examples of every degree, however, e.g. $(a+ib-c-id)^4$ is invariant under the transformation $a\mapsto b\mapsto c\mapsto d\mapsto a$.
