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This makes sense to me if $a_j\ne a_k$ for $j\ne k$ as $(x-a_j)=0 \implies a_j$ is a root of $f(x)$. So if all $a_j$ are different, then all the roots will be different. Do I have to somehow show this using the fact that gcd$(f(x),f'(x))=1$, because if so, I don't know how to

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  • $\begingroup$ You have to show two things: 1) that if all the $a_i$-s are distinct, then $\operatorname{gcd}(f,f')=1$; 2) that if $a_i=a_j$ for some $i\ne j$, then $\operatorname{gcd}(f,f')\ne 1$. $\endgroup$ – Saucy O'Path May 20 at 20:39
  • $\begingroup$ Hint: can you write an expression for $f'(x)$ given the above representation for $f(x)$? $\endgroup$ – user293794 May 20 at 20:40
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Compute the derivative: $$f'(x)=(x-a_2)\cdots(x-a_n)+(x-a_1)(x-a_3)\cdots(x-a_n)+\ldots+(x-a_1)\cdots(x-a_{n-1})$$ Suppose $f$ has a multiple root; wlog assume $a_1=a_2$. Then in the sum above, each term is divisible by at least one of $(x-a_1)$ and $(x-a_2)$. Hence $f'(a_1)=0$, and thus $(x-a_1)\mid\mathrm{gcd}(f,f')$.

Conversely, suppose $f$ has no multiple roots. Then $$f'(a_i)=(a_i-a_1)\cdots(a_i-a_{i-1})(a_i-a_{i+1})\cdots(a_i-a_n)$$ Since $f$ has no multiple roots, all these factors $(a_j-a_i)$ are nonzero. Hence $f'(a_i)\ne0$, thus no factor $(x-a_i)$ of $f$ divides $f'$, which means that the gcd is $1$.

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