> 
> Suppose that polynomial $x^4+x+1$ has multiple roots over a field of
>  characteristic $p$ . What are the possible values of $p$?
> 
> 
> 

My solution : 
Set $f=x^4+x+1$. Suppose the multiple root is $m$ . So $f,f'$ (the formal derivative of $f$) have root $m$ in the field of characteristic $p$. 

Hence $m^4+m+1=0 \pmod{p} $ and $4m^3+1=0 \pmod{p}$ 

So $3m+4=4(m^4+m+1)-m(4m^3+1)=0 \pmod{p}$ .

From here it is easy to to see that $p\neq 3$. I have no idea if it is possible to proceed any further. Please provide a solution.
['https://math.stackexchange.com/questions/2235118', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/71612/']
Hint:

Find the quotient and remainder when $(27)(4m^3 + 1)$ is divided by $3m+4$.
Using the [extended Euclidean algorithm](http://www.wolframalpha.com/input/?i=PolynomialExtendedGCD%5Bx%5E4+%2B+x+%2B+1,+4+x%5E3+%2B+1%5D) we get
$$
229 = (x^4 + x + 1)(144 x^2 - 192 x + 256)+(4 x^3 + 1)(-36 x^3 + 48 x^2 - 64 x - 27)
$$
This can also be found by computing the [resultant](http://www.wolframalpha.com/input/?i=resutant%5Bx%5E4+%2B+x+%2B+1,+4+x%5E3+%2B+1%5D) of $x^4 + x + 1$ and $4 x^3 + 1$, that is, the discriminant of $x^4 + x + 1$.