From Romania TST 2004 Day 5 P3, I was introduced to the polynomial $$f(x)=\sum_{i=1}^{p-1} \left( \frac{i}{p} \right)x^{i-1}$$ This polynomial is clearly not irreducible - $x=1$ is a root.
Even more - this MO thread gives that for $p \equiv 1 \pmod{4}$, we have $$f(x)=(x-1)^2(x+1)k(x)$$ for a integer polynomial $k(x)$.
While the MO thread above asks for irreducibility of the $k(x)$, I am interested in the irreducibility of $$g(x)=1+xf(x)$$
Now I checked that this is irreducible for $p \le 17$ (I know, not a lot).
I tried to use Eisenstein Criterion to prove it - just like the irreducibility of cyclotomic polynomials, I substituted $x+1$ instead of $x$, but then I get $$g(x)=1+ \sum_{i=1}^{p-1} \left( \frac{i}{p} \right) (x+1)^i$$
Now let $$g(x) = \sum_{i=0}^{p-1} a_ix^i$$
We get $$a_k = \sum_{i=k}^{p-1} \binom{i}{k} \cdot \left(\frac{i}{p}\right)$$ So $a_{p-1} = \left( \frac{-1}{p} \right)$ and $a_0=1$ which is okay, but I can't seem to find a prime $q$ such that $q|a_i$ for all $1 \le i \le p-2$, i.e. an appropriate prime to use the Eisenstein Criterion.
Checking again for $p \le 11$, I couldn't find a prime for all of them.
So I scratched the $x+1$ idea and moved on with $x-1$.
We would get $$a_k = \sum_{i=k}^{p-1} (-1)^{i-k}\binom{i}{k} \cdot \left(\frac{i}{p}\right)$$
The problem is that the right primes for Eisenstein does not appear here as well.
I've tried to use Perron's and Cohn's Criterion but it obviously fails.
Question. Is $g(x)$ irreducible for all $p$?
If not, what is the counterexample?
If correct, is there a suitable irreducibility criterion for this problem?
Am I using 'overpowered' techniques on a problem that is easy? (I'm pretty sure I'm not)