Elementary methods for checking double roots How can I check if a large polynomial has any double roots? I was trying to see if the polynomial
$$x^{2021} + x^{3} + 1$$
had any double roots, but I had no idea what to do. After some research, I saw that that the polynomial should never be equal to it's derivative, but unfortunately I am unfamiliar with calculus. I tried seeing if I could find some roots and perhaps discover anything interesting, but it would take too long.
Is there an elementary method of verifying this?
 A: The powers $2021$ and $3$ are both odd, hence the terms $x^{2021}$ and $x^3$ are both monotonically increasing, hence so is their sum.  Adding $1$ merely shifts the sum up a bit.  No double roots.
Incidentally, the test for multiple roots is not 'is the polynomial equal to its own derivative' (never true for polynomials), it is 'does the polynomial share a factor with its own derivative'.
A: I shall admi that we are in the real domain
$$f(x)=x^{2021}+x^3+1 \implies f'(x)=x^2 \left(2021 x^{2018}+3\right)$$ $x=0$ is not a root of $f(x)$ and the next term is always positive.
This case is easy but, in a general manner, ???
A: A followup to the answer by Claude Leibovici.
This is a quick way of seeing it, but it might not be elementary enough:
By the Eisenstein criterion the polynomial $2021x^{2018}+3$ is irreducible over $\Bbb Q$. Hence $\gcd(f,f')\ne1$ if and only if $2021x^{2018}+3\mid f$. But this is impossible as $x^{2018}+\frac{3}{2021}\notin\Bbb Z[x]$.
(In general if $R$ is a UFD and $f\in R[x]$ monic with $f=gh$, $g,h\in (\operatorname{Frac}(R))[x]$ monic then $g,h\in R[x]$)
Therefore $\gcd(f,f')=1$, i.e. $f$ has no multiple roots.
A: You may apply the classic techniques in formal derivatives whose definition doesn't rely on calculus.  Let $f(x) = x^{2021} + x^3 + 1$.  Then $f'(x) = 2021x^{2020} + 3x^2 = (2021x^{2018}+3) x^2$, which has

*

*a double root $x = 0$.

*$2018$ $4036$-th roots of unity $\omega_{4036}^{2k+1}$ with $k \in \{0, 1, \dots, 2017\}$ multiplied by $\sqrt[2018]{3/2021}$.

It suffices to verify that these $2019$ roots of $f'(x)$ aren't a zero of $f(x)$.
\begin{aligned}
&f(\omega_{4036}^{2k+1}) \\
=& (\omega_{4036}^{2k+1})^3 (-3/2021) + (\omega_{4036}^{2k+1})^3 + 1 \\
=& 1+2018\omega_{4036}^{3({2k+1})}/2021
\end{aligned}
Note that $\mathrm{Im}(f(\omega_{4036}^{2k+1})) = 2018 \sin(3(2k+1)\pi/2018)/2021 \ne 0$ as $3(2k+1)/2018 \notin \mathbb{Z}$ for all $k \in \mathbb{Z}$.
