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Take an odd prime $p$ and put $x_0:=\sum\limits_{j=0}^{p-1}\left(a_{j}\sqrt{p}\cos\dfrac{j\pi}p+b_{j}\sin\dfrac{j\pi}p +c_{j}\tan\dfrac{j\pi}p\right)$, where the $a_{ij}$ are integers. If $f$ denotes the minimal polynomial of $x_0$, can we prove that $p$ divides all coefficients of $f$ except the leading one?

I have quite a bit of numerical evidence for this. Note that it obviously doesn't hold without the $\sqrt p$ factor, but more interestingly, it is also false if $\sqrt p$ goes with the other terms instead of the $\cos$ term. Moreover, it seems that in those cases, none of the non-zero coefficients is divisible by $p$.

(More generally, I think those coefficients are divisible by $p$ if we replace $\dfrac{j\pi}p$ by $\dfrac{j\pi}{p^r},\ r\in\mathbb N$ and do the sum over $j=0,...,p^r-1$.)

If all but one of the $a_j,b_j,c_j$ are $0$, the claim is quite easy to prove (and not new). For instance, for $x_0=\sin\dfrac{j\pi}p$ with any fixed $j$, we have explicitly $$f(x)=\sum\limits_{i=0}^k(-1)^k\dbinom p{2i+1}(1-x^2)^{k-i}x^{2i},$$ where $p=2k+1$. So the claim is obvious here.

Added: It should be clear from Galois theory that in general, the conjugates of $x$ are the sums obtained by replacing all the $j$'s by $kj$ for a fixed $k=2,...,p-1$.

Literature:

Beslin, S., de Angelis, V., 2004. The minimal polynomials of sin(2π/p) and cos(2π/p). Mathematical Magazine 77, 146–149.

Heierman, William E., Minimal polynomials for trig functions of angles rationally commensurate with π

Lang, Wolfdieter, Minimal Polynomials of sin (2π/n)

Surowski, David, and McCombs, Paul, Homogenous polynomials and the minimal polynomial of cos(2π/n)

W. Watkins and J. Zeitlin, The Minimal Polynomial of cos(2Pi/n), Am. Math. Monthly 100,5 (1993) 471-4.

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I can't quite do this all in my head, but clearly enough you should should write 1 - ζ = π and do π-adic analysis in the cyclotomic field of p-th roots of unity, where ζ is a non-trivial p-th root of 1. Putting the factor in front of the cos term means it doesn't affect the early terms of the expansion. (To be safe you have to put the square root of -1 in the field also.) The rest I presume can be read off the Newton polygon somehow, perhaps with a bit more work.

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I have just edited the expression for $x$ to make it (slightly) more general in a quite obvious way, and I think what you suggest should still apply. Only I don't really understand what you presume, particularly what do you mean by "early terms of the expansion"? –  Wolfgang Jan 25 '12 at 18:34
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A π-adic expansion works just like a Taylor expansion, and here the sin and tan terms are of order 1, while the square root of p is of order (p - 1)/2. –  Charles Matthews Jan 25 '12 at 19:36
    
As Charles pointed out, consider $\mathbb{Q}(\zeta_p, i)$, and a prime ideal $\mathcal{P}\subset \mathbb{Q}(\zeta_p,i)$ lying over $(1-\zeta_p)\subset \mathbb{Q}(\zeta_p)$. Then, $\sqrt{p}$, sin, and tan terms are divisible by $\mathcal{P}$ considered as in $\mathcal{P}$-adic integer, while cos term alone is not. Putting $\sqrt{p}$ on other terms will leave cos term alone, so this will not be divisible by $\mathcal{P}$. Anyway, very interesting observation. –  i707107 Mar 8 '12 at 4:37
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