> 
> 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.
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Hint:

Find the quotient and remainder when $(27)(4m^3 + 1)$ is divided by $3m+4$.
$$3m+4 \equiv 0 \pmod{p} \\
3m \equiv -4 \pmod{p} \\
27m^3 \equiv -64 \pmod{p}$$

You also have 
$$4m^3\equiv -1 \pmod{p}$$

Denote $m^3=:x$ then
$$27x \equiv -64 \pmod{p} \\
4x \equiv -1 \pmod{p}$$

Multiply first equation by 4, second by 27 and subtract.