# Generating polynomials that are co-prime to their first and second derivatives

Let $f \in \mathbb Q [X]$ and not constant or of the form $(x-a)^n$. Suppose:

$f_1 := \frac{f}{gcd(f,D^2f)}$ and;

$f_2 := \frac{f_1}{gcd(f_1,Df_1)}$,

where $Df$ stands for the formal derivative.

Is it true that $gcd(f_2,Df_2)=gcd(f_2,D^2f_2)=1$

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Yes, it is true. –  Eric Naslund Apr 12 '11 at 18:45
@ Eric ..Can you suggest proof of the statements. –  Ram Apr 12 '11 at 19:01
@Ram: are you and the other one the same Ram? If so you seem to have been logging in using to different Google IDs, which resulted in your two accounts disassociating. E-mail me and I'll merge them. –  Willie Wong Apr 12 '11 at 19:06
If I'm not mistaken, you get $\gcd(f_2,Df_2) = 1$, but the difficulties come with the second derivative. It is easy to characterize polynomials that are relatively prime to their derivative, but relatively prime to their second derivative is not so immediate. –  Arturo Magidin Apr 12 '11 at 19:32

## 1 Answer

No, it is not true. Try $f(x) = x^2 + x^4$

$f_1(x) = x^2 + x^4$

$f_2(x) = x + x^3$

${\rm gcd}(f_2, D^2 f_2) = x$

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