# 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?

You may apply the classic techniques in formal derivatives whose definition doesn't rely on . 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 $$k \in \mathbb{Z}$$.

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'.

• Thanks for the help? I had thought about a graphical approach, but I didn't follow through. Thanks so much! – WWesEEE Jan 13 at 13:51
• I suppose one might say that at a double root the polynomial and its derivative are both zero, so at that point the polynomial and its derivative are equal--but I think we would agree that this is not a particularly useful way to explain anything. – David K Jan 13 at 13:55

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.

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, ???