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

Question

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.

Answer 1

Hint:

Find the quotient and remainder when $(27)(4m^3 + 1)$ is divided by $3m+4$.

Answer 2

Using the extended Euclidean algorithm 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 of $x^4 + x + 1$ and $4 x^3 + 1$, that is, the discriminant of $x^4 + x + 1$.

Answer 3

$$3m+4 \equiv 0 \pmod{p} \\ 3m \equiv -4 \pmod{p} \\ 27m^3 \equiv -64 \pmod{p}$$

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.