# Why is the sum of the ($\varphi(d)$-1)th coefficients of $\Phi_d$ equal to the ($\varphi$(n)-1)th coefficient of $\prod_{d\mid n} \Phi_d(X)$

I grasp this answer, except for one identity. To quote: "$\sum_{d\mid n}\left[\Phi_d(X)\right]_{-1} = \left[\prod_{d\mid n} \Phi_d(X)\right]_{-1}$"

It isn't so simple i think because you don't take the coefficients corresponding to the same place/x-power. What is my misunderstanding?

Update: I simply had missed reading the updates, which made it all reasonable. Thanks to https://math.stackexchange.com/a/69548/218659 I understand that "the coefficient in question is the negative sum of the roots".

Though I feel somewhat silly, i honestly can't produce myself why "Even without considering roots, this follows from looking at how the product expands."

• See the updates in that answer. Then apply induction to get this for product of any number of polynomials. – Wojowu Dec 21 '15 at 14:30
• updated my question – Grzegorz Baltissen Dec 31 '15 at 12:10

Given two monic polynomials $f_1=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0,f_2=x^m+b_{m-1}x^{m-1}+...+b_1x+b_0$, after we multiply them we will get a polynomial of degree $n+m$. After grouping terms containing $x^{n+m-1}$ in the product, we notice we get $1\cdot b_{m-1}+a_{n-1}\cdot 1=a_{n-1}+b_{m-1}$. This is precisely the statement that next-to-leading coefficient of the product of two monic polynomials is the sum of their next-to-leading coefficients. Now, by induction, this is true for any number of polynomials in a product. After applying this to the cyclotomic polynomials and their product we get what was asked for.