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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?

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  • $\begingroup$ 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. $\endgroup$ Nov 19, 2017 at 3:01
  • $\begingroup$ 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 $\endgroup$ Nov 19, 2017 at 3:07

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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.

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  • $\begingroup$ Why this is true? $$ \gcd(f(x),f'(x)) = \gcd(f'(x), n f(x)-x f'(x)) $$ $\endgroup$
    – IrbidMath
    Nov 19, 2017 at 4:33
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    $\begingroup$ @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. $\endgroup$ Nov 19, 2017 at 4:37
  • $\begingroup$ We can prove it by induction on the degree i guess? $\endgroup$
    – IrbidMath
    Nov 19, 2017 at 4:40
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    $\begingroup$ @Ameryr: no induction is needed. In a Euclidean ring, if $d$ is a divisor of both $a$ and $b$, $d$ is a divisor of $a\cdot\text{whatever}\pm b\cdot\text{whatever}$. $\endgroup$ Nov 19, 2017 at 4:44
  • $\begingroup$ Oh. Yeah thats true any linear combination $\endgroup$
    – IrbidMath
    Nov 19, 2017 at 4:47

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