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Let A be the set of all polynomials f with positive integral coefficients such that $f(n)|(2^n-1)$ for all $A\in\mathbb{N}$. Then $$\sum_{f\in A} {1\over f(2019)}=?$$

I tried to define a polynomial, but where is the degree? Hence I considered all polynomials that leave remainder $1$ mod $f.$ Hence $2^n-1$ is divisible. But actually I cannot guess how to proceed! Please help!

Lee David Chung Lin
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  • What is the source of all the questions you are asking? – lulu Jan 20 '19 at 15:36
  • My unsolved problems in todays exam –  Jan 20 '19 at 15:36
  • Do you mean ... Let $A$ be the set of all polynomials $f$ with positive integral coefficients such that $f(n)|(2^n-1)$ for all $n \in \mathbb{N}$. Then $$\sum_{f\in A} \frac{1}{f(2019)}=\text{?}$$ (Note that, along with formatting, I replaced "for all $A\in\mathbb{N}$" with "for all $n\in\mathbb{N}$".) – Blue Jan 20 '19 at 15:45
  • Yes! But i could not type it so replaced it –  Jan 20 '19 at 15:51
  • Please use MathJax to format your questions. You can take a look at how I edited your post, to begin with. – saulspatz Jan 20 '19 at 15:55
  • Can someone help please? –  Jan 20 '19 at 15:56
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    Let us start by trying to understand the set $A$ better. According to the conditions on $f$ we must have $f(1) = 1$ (take $n=1$ and use that $f$ has positive coefficents). But since $f$ has positive, integral coefficients, this seems to indicate that $f$ must have the form $f(x) = x^k$. But in order for an $f$ on this form to meet the condition we would have to have that $k$ should divide $2^k -1$ for all $k$, which is clearly not true. So I suspect $A$ is empty, hence the sum is 0. – Testcase Jan 20 '19 at 17:29
  • @Testcase: Don't forget $f(x)\equiv 1$. (BTW: You can eliminate the possibility of $f(x)=x^k$ for $k\geq 1$ by simply noting that $2^k \not\mid 2^2-1$.) – Blue Jan 20 '19 at 18:22

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