For natural numbers $n$ and $x,$ the number of $n^{th}$ roots that have $x$ in the whole numbers place can be represented as $(x+1)^{n}-x^{n}.$ For $p$ prime,
$(x+1)^{n}-x^{n}-1\equiv0\bmod p$ iff $n=p^{k}.$
The AKS test uses this identity at $k=1$ giving,
$(x+1)^{n}-x^{n}-1\equiv0\bmod n$ iff $n$ is prime.
I have found that it is also interesting if we evaluate the exponent at only the even numbers $2n,$ meaning in this form,
$(x+1)^{2n}-x^{2n}-1.$
In this form, the smallest number $m$ that will not divide each and every coefficient of the expansion seems to be the greatest prime factor of $2n+1.$ In other words, If you choose $2n=32$ then dividing by $2,3,4,5,6,7,8,9$and $10$ will give some integer coefficients, but when you divide by the greatest prime factor of $33$ which is $11$, there are none. This is also the smallest number for which that occurs. So this can be used as a test.
Now to determine what this greatest prime factor is it seems that you would have to check every term, or at least half of them, so I'm not sure if it can be afforded the same computation simplification as the AKS test, but this still may be a polynomial time test for greatest prime factor of $2n+1.$
So my question is, does this idea afford a polynomial time test for this or is it a useless idea?
To my best knowledge, this is unknown. But, I of course could be wrong.
It seems this question was answered in the comments, so I'm using an answer to my own question to close the question. If anyone has an answer to give they are welcome to do so and I will probably accept it.