<s>[INST] Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 = 1/4 ((2x+1)^2+ 3)$$
$$ x^4 + x^3 + x^2 + x + 1 = 1/16((4 x^2 + 2 x + 1)^2 + (2 x + 3)^2 + 6)$$
$$ x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 1/144 ((12x^3+6x^2+4x+3)^2+ 3(2x^2+2x+5)^2 + 5(2x+3)^2 +15)$$
 [/INST] $$
P(x) = \prod_{d \mid 2n+1, d \neq 1} \Phi_d (x)\\
$$
So focus on getting a good decomposition for $\Phi_d (x)$.
$\Phi_d (x)$ is a product of $\phi(d)$ terms. $d$ is not even because it divides $2n+1$ and it is greater than $1$ by assumption. For the factor corresponding to $k$ there is also the factor corresponding to $d-k$ and those two are conjugate and distinct so $\Phi_d (x) = Q(x) \overline{Q}(x)$ for some $Q$ as before but with the caveat that now we are taking only other primitive roots of unity rather than all powers of $\omega$. But that doesn't really help because the decomposition is still using polynomials with irrational and hard to compute coefficients.
Another property $\Phi_n (x) = \Phi_q (x^{n/q})$ where $q$ is the radical of $n$ by only keeping the same prime factors but only with multiplicity $1$.
So we focus on providing good sum of squares decomposition for $\Phi_q (x)$ where $q$ is a square free number dividing $2n+1$ and greater than $1$.
If $q=3$, then
$$
\Phi_3 (x) = x^2 + x + 1\\
= (x+\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2\\
$$
Otherwise Gauss' formula applies
$$
\Phi_q (x) = (\frac{1}{2} A_n (x))^2 - (-1)^{(q-1)/2} q (\frac{1}{2} x B_n(x))^2\\
$$
When $q \equiv 3 (4)$ this gives a decomposition into a sum of $q+1$ squares of polynomials. All the coefficients involved are half-integers in this case.
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