Factor the polynomial $f(x)=x^{15}+3x^{10}+2x^5+4$ in $\mathbb{F}_5[x]$.
Alright, so I'm sure this is a simple answer but I have not been able to figure out, hopefully I can quickly be set on the right track.
Typically to do this I would find $f'(x)$, then $\gcd(f,f')$ and proceed from there vis-a-vis the Square-Free method of factoring polynomials over finite fields.
However, $f'(x)=0$ in $\mathbb{F}_5[x]$. I did not see any reference to this case in my textbook, but I did find the following explanation on Wikipedia, which seems like it should be useful here:
[The Square-free] algorithm uses the fact that, if the derivative of a polynomial is zero, then it is a polynomial in $x^p$, which is, if the coefficients belong to $\mathbb{F}_p$, the $p^{th}$ power of the polynomial obtained by substituting $x$ by $x^{1/p}$.
What is meant here by substituting $x$ by $x^{1/p}$?
My first thought was to substitute each term in the polynomial, e.g. $x^{15} \rightarrow \left(x^{15}\right)^{1/5}=x^3$. But I don't know how that substitution would apply to $4$, since $4^{1/5}$ doesn't make much sense in $\mathbb{F}_5$.
Consider the polynomial $$ 4+2 \left(x^{1/5}\right)^5+3 \left(x^{1/5}\right)^{10}+\left(x^{1/5}\right)^{15} =x^3+3 x^2+2 x+4\, $$ In a field of characteristic $p$, we have that $$ (a+b)^p = a^p +b^p\,, $$ so that, by a quick finite induction argument, $$ (x^3+3 x^2+2 x+4)^5 = \left(x^3\right)^5 + \left(3x^2\right)^5 + \left(2x\right)^5 + 4^5 = x^{15}+3^5x^{10}+2^5x^5+4^5x^5\, $$ which reduces to the original polynomial by Fermat's little theorem.
If we had forgotten about the above rule (like me, earlier), we can just compute the fifth power of this polynomial using Mathematica, yielding $$ x^{15}+15 x^{14}+100 x^{13}+410 x^{12}+1225 x^{11}+2923 x^{10}+5730 x^9+9420 x^8+13280 x^7+15920 x^6+16352 x^5+14400 x^4+10240 x^3+6400 x^2+2560 x+1024\, $$ which can be seen to reduce to $$ x^{15}+3 x^{10}+2 x^5+4 $$ mod 5.