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Given the nature of the exercise, I think the point is not using the Fundamental Theorem of Algebra to show there must be $n$ roots. Anyway, the polynomial is the following: $$x^n + x + 1$$

So far I've tried to reach an absurd by proposing a root that is both root of the polynomial and its derivative, but I'm stuck there. Any hints for this kind of exercise?

Francisco José Letterio
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  • Yes, you should consider the case that the polynomial has a common root with its derivative. That means that the polynomial and its derivative have a common divisor. You can find the greatest common divisor (gcd) using the Euclidean Algorithm. That is a hint, with that you will be able to solve the problem. – Zach Teitler Nov 19 '17 at 03:01
  • Im entering a loop here. I get (x^n +x+1:nx^(n-1) + 1) = (nx^(n-1) + 1:(n-1)x/n + 1) and when I start dividing this I get stuck in a loop – Francisco José Letterio Nov 19 '17 at 03:07

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We may safely assume $n\geq 2$. If $f(x)=x^n+x+1$ we have $f'(x)=n x^{n-1}+1$ and $$ \gcd(f(x),f'(x)) = \gcd(f'(x), n f(x)-x f'(x))=\gcd(nx^{n-1}+1,(n-1)x+n). $$ On the other hand $(n-1)x+n$ only vanishes at $x=\frac{n}{1-n}$, while $$ n\left(\frac{n}{1-n}\right)^{n-1}+1 $$ is positive for any odd $n$ and strictly negative for any even $n$, since $n^n>(n-1)^{n-1}$.
It follows that $f(x)$ and $f'(x)$ cannot have common roots, hence all the roots of $f(x)$ are simple.

Jack D'Aurizio
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  • Why this is true? $$ \gcd(f(x),f'(x)) = \gcd(f'(x), n f(x)-x f'(x)) $$ – IrbidMath Nov 19 '17 at 04:33
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    @Ameryr: any polynomial which simultaneously divides both $f(x)$ and $f'(x)$ also simultaneously divides both $f'(x)$ and $nf(x)-x,f'(x)$. In particular the greatest-common-divisor of $f(x)$ and $f'(x)$ is also the greatest-common-divisor of $f'(x)$ and $n f(x)-x f'(x)$. It is just the main principle of the Euclidean algorithm. – Jack D'Aurizio Nov 19 '17 at 04:37
  • We can prove it by induction on the degree i guess? – IrbidMath Nov 19 '17 at 04:40
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    @Ameryr: no induction is needed. In a Euclidean ring, if $d$ is a divisor of both $a$ and $b$, $d$ is a divisor of $a\cdot\text{whatever}\pm b\cdot\text{whatever}$. – Jack D'Aurizio Nov 19 '17 at 04:44
  • Oh. Yeah thats true any linear combination – IrbidMath Nov 19 '17 at 04:47
  • @JackD'Aurizio: by your logic, if 2 divides 4 and 8, then 2 divides 4 * 1 - 8 * 0.125 = 3... – user541686 Nov 19 '17 at 11:34
  • @Mehrdad: when $a,b$ and $\text{whatever}$ belong to the same integral domain that is certainly true, so yours is not a counter-example. Again, just see the Euclidean algorithm for polynomial division. – Jack D'Aurizio Nov 19 '17 at 17:31
  • @JackD'Aurizio: But my point was $x$ is not necessarily an integer (we have $x f'(x)$)... right? Note that I'm not saying the conclusion is wrong, I'm talking about the reasoning your just provided. – user541686 Nov 19 '17 at 19:38
  • @Mehrdad: of course the implementation of the Euclidean algorithm depends on which Euclidean domain we are considering. If we work in $\mathbb{Z}$, we say that $d\mid a, d\mid b\to d\mid(ka-jb)$ for any $a,b,d,j,k\in\mathbb{Z}$. If we work in $\mathbb{Z}[x]$, we state the same by meaning that $a,b,d,j,k\in\mathbb{Z}[x]$. – Jack D'Aurizio Nov 19 '17 at 19:44
  • @JackD'Aurizio: Yeah. All I'm trying to say is that that's far less trivial or obvious of a statement than you made it seem in your comment above... i.e., I think you over-simplified your comment. :) Polynomial division is taught in high school, but $\mathbb{Z}[x]$ is something non-mathematicians generally don't ever encounter. So chances are someone looking at your comment about $d | a \cdot \text{whatever} \pm b \cdot \text{whatever}$ would interpret it naively (such as the interpretation I gave you above) and entirely miss the crux of it. – user541686 Nov 19 '17 at 20:20
  • I'm still struggling a little with this. Why do you claim it is the greatest common divisor of f' and that linear combination? I know that it divides it, but not why it must be the GCD – Francisco José Letterio Nov 21 '17 at 18:58
  • @FranciscoJoséLetterio: again, every common divisor to both $a$ and $b$ is also a common divisor to both $b$ and $a-qb$. In particular, the greatest common divisor of $a,b$ equals the greatest common divisor of $b,a-qb$. This is the foundation of Euclid's algorithm. – Jack D'Aurizio Nov 21 '17 at 19:16
  • But what about, for example, $(x +1,x-1) and (x+1, x+1 + (x^2 +2x +1)(x-1)$. Clearly not the same GCD – Francisco José Letterio Nov 21 '17 at 19:20
  • @FranciscoJoséLetterio: Sorry, I took a bad formulation (now fixed). Let us say that an allowed reduction is $(a,b)\to (b,a-pb)$. By this way the $\gcd$ is preserved and clearly the multiplication by $n$ of some term does not affect the $\gcd$ if we are talking about polynomials in $\mathbb{R}[x],\mathbb{Q}[x]$ or $\mathbb{C}[x]$. – Jack D'Aurizio Nov 21 '17 at 19:30
  • I am sorry but I don't know what a reduction is. Thanks anyway – Francisco José Letterio Nov 21 '17 at 20:19