1) If $f(x)=\frac{x^n}{n!}+\frac{x^(n-1)}{(n-1)!}...x+1$ then prove that f cannot have repeated roots.

This one I did not expect it:

2) If f(x) is a polynomial with integer coefficients such that f(0)=1 and f(1)=5, then prove that f cannot have integer roots.

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## Comments

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TopNewestFor 1., you can use the fact that if $f(x)$ has a repeated root, then $f'(x)$ has the same root, say $a$ .

This implies that $f(a) - f'(a) = 0 \implies \dfrac{a^{n-1}}{(n-1)!} = 0 \implies a = 0$. But $0$ is obviously not a root of $f(x)$.Therefore, there exists no repeated root of $f(x)$.

For 2, you can see that modulo 2, $f(a) \equiv f(1) , f(0) \equiv 1$( depending on whether a is odd or even). for all integral $a$. But, if a integral root $b$ did exist, that would imply that $f(b) = 0 \equiv 0$, which leads to a contradiction

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Ya. I used the exact same methods for both the questions. Great minds think alike ; )

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Let us assume that $f_n(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}$ has no repeated roots.

We now study the function $f_{n+1}(x)$

$f_{n+1}(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}$

Observe the derivative of $f_{n+1}(x)$ $f_{n+1}^{\prime}(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}$

Clearly, $f_{n+1}^{\prime}(x)$ has no repeated roots and must have $n$ distinct roots by our inductive hypothesis. Therefore, by Rolle's Theorem, we must have $n+1$ distinct roots of $f_{n+1}(x)$. And thus proved.

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@arian tashakkor

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ok these ones are somewhat hard to understand I should say ...

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I've posted some more proof notes.

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The first one can be done with the help of LMV or Rolles i guess??

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The first one is pretty simple. I don't like using LMV or Rolle's theorem, or at least their terms in anything that can be solved with a rough sketch. Although I don't use it, my method is pretty much related to it. If the function has repeated roots, then it's derivative also has a root at that point. If that's the case, then the $x^n/n!$ factor in the function has no consequence in altering the value taken by the function. In that case, x must have a repeated root at 0. But substituting 0 in the place of x gives you 1. So 0 is not a root and the function doesn't have any repeated roots.

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