Positive coefficients after the substitution? Let $n>1$ be an integer and $\Phi_n(x)$ be the $n$ th cyclotomic polynomial.

If we subsititute $x$ by $x+1$ , do we always get a polynomial with positive coefficients ? If yes, how can this be proven ? I also want to prove that no
coeffcient upto the $\varphi(n)$ th is missing.

Examples :

*

*$\Phi(6)=x^2-x+1$ transforms into $x^2+x+1$

*$\Phi(24)=x^8-x^4+1$ transforms into $x^8+8x^7+28x^6+56x^5+69x^4+52x^3+22x^2+4x+1$
Upto $n=3\ 200$ , this is the case. Some special cases can be easily proven, for example that $n$ is a power of $2$ or that $n$ is a prime. But I have no idea for a general proof.

Deal with the case $n=1,$ $ n = 2$ separately.
Notice that $ \Phi_1(x+1) = (x+1) - 1 = x$, which has a non-positive constant term.
This will be the only coefficient of $ \Phi_n (x+1)$ which is non-positive.

For $ n > 2$,
Let $A_n$ denote the set of integers $ 0 < k < n$ such that $ \gcd( n,k) = 1$.
Let $A_n^*$ denote the set of integers $ 0 < k < n/2$ such that $ \gcd( n,k) = 1$.
$$\Phi_n(x)  = \prod_{j \in A_n } (x - e^{2k\pi/n}) =  \prod_{j \in A_n^*} \left( x^2 - 2 \cos \frac{ 2\pi j}{n} \cdot x + 1 \right) .$$
Let $ f_m (x)  =  x^2 - 2mx + 1 $.
Since $f_m(x+1) = x^2 + 2(1-m)x + 2 (1-m)$, so the coefficients of $f_m(x+1)$ are positive for $ m < 1$.
Thus, for $n > 2$, since $ \cos \frac{2 \pi j } { n} < 1$ for $ j \in A_n^*$
$$\Phi_n(x+1) = \prod_{j \in A_n^*} f_{\cos \frac{2\pi j  }{n}}(x+1) ,$$
where the latter is a product of polynomials with positive coefficients, hence $\Phi_n(x+1)$ has positive coefficients for up to the degree $ \phi(n)$.
