1

I found the following Math Stack Exchange article, which proves this statement for N = 2015 but the subset of elements should be divisible by 5 and I am trying to understand the first answer of the following article.

Find the number of all subsets of $\{1, 2, \ldots,2015\}$ with $n$ elements such that the sum of the elements in the subset is divisible by 5

There are two things I don´t understand about it.

  1. I don´t get why the last equality in the equation (*) holds. I get that the scalar before $x^{|S|}$ is equal to 1 if $\sum S = 0 \text{ mod } 5 $. But I don´t understand what happens to the scalars w.r.t to subsets, which are not divisible by 5.

  2. I don´t understand how $$ f(\omega^j,x) = (1+x^5)^{403} \text{ for } j = 1,2,3,4 $$ can be derived. I am looking for a calculation proving that.

In general I am aware, that there is an article, proving my question for prime numbers, but I am particularly interested, in the case where n is not necessarily prime.

Unfortunately my reputation is too low to comment the original post :/

Desperate_housewife
  • 123
  • 2
  • If $\Sigma S\not\equiv 0 \pmod{5}$, then we're adding the five distinct 5th roots of unity when we sum up $\sum_{j = 0}^4 (\omega^{\Sigma S})^j$. 2. Kind of the same idea; the terms in the product repeat every 5th $k$, so you just need to compute $\prod_{k = 1}^5 (1 + \omega^k x)$.
    • – Jakob Streipel Nov 13 '24 at 14:50
  • @JakobStreipel Thanks for the comment. 1. Thanks! 2. I did that, but I don´t understand why one can omit the potence k in every factor and consequently why the product you have written is equal to $1+x^5$ – Desperate_housewife Nov 13 '24 at 15:26
  • 1
    @Desperate_housewife You can tediously expand out $ ( 1 + \omega x)( 1 + \omega^2 x)( 1 + \omega^3 x)( 1 + \omega^4 x)( 1 + \omega^5 x)$ and verify that it is equal to $x^5 + 1$. Use $ \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0 $. EG The $x^5$ term has coefficient $\omega^{15} = 1$, the $x^4$ term has coefficient of $\omega^4 + \omega^3 + \omega^2 + \omega + 1$, which is 0. $\quad$ For a smart derivation, observe that the roots of $x^5 + 1$ are $ - \omega^k$, so $ x^5 + 1 = \prod ( x + \omega^k) = \prod ( \omega^{5-k} x + 1 )$. – Calvin Lin Nov 13 '24 at 22:12
  • @CalvinLin Thanks that part was very helpful and makes a lot of sense now. But why is $\prod_{k=1}^{5} (1 + \omega^{kj}) = \prod_{k=1}^{5} (1+ \omega^k)$ for j = 1,2,3,4. As this step occurs in the calculation of $f(\omega^j,x)$ ? – Desperate_housewife Nov 14 '24 at 14:37
  • Just write it out. Show that $ { j, 2j, 3j, 4j } = { 1, 2, 3, 4 }$ for $j = 1, 2, 3, 4$, which should eventually obvious. – Calvin Lin Nov 14 '24 at 16:14