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One can show that

$$\frac{(X^{15} - 1)(X-1)}{(X^5-1)(X^3-1)} = P^2 + 15 Q^2$$ with $P,Q\in\mathbf{Z}\left[\frac{1}{2}\right][X]$, by simply grouping terms etc. Is it possible to show that such an identity exists by arguments involving $\mathbf{Z}\left[\sqrt{15}\right]$ or other rings over the latter ?

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  • $\begingroup$ This is impossible in $\mathbb{Z}[1/2]$. It took me 3 hours to get me no were. $\endgroup$ – Aqua Nov 4 '17 at 19:14
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Nice question! The LHS clearly is the cyclotomic polynomial $\Phi_{15}(x)$, and by considering its factorization over the ring of integers of $\mathbb{Z}[\sqrt{5}]$ we have

$$ \Phi_{15}(x) = \frac{1}{4}\left(2+\left(-1+\sqrt{5}\right) x+\left(1-\sqrt{5}\right) x^2+\left(-1+\sqrt{5}\right) x^3+2 x^4\right)\cdot\left(2-\left(1+\sqrt{5}\right) x-\left(-1-\sqrt{5}\right) x^2-\left(1+\sqrt{5}\right) x^3+2 x^4\right) $$ so $$ \Phi_{15}(x)=\frac{1}{4}\left[(1+x+x^2)^2(2x^2-3x+2)^2-5x^2(x^2-x+1)^2\right]$$ and a similar decomposition follows from considering how $\Phi_{15}$ factors over the ring of integers of $\mathbb{Z}[\sqrt{5},\sqrt{-3}]$. One factor is namely given by $$\left(-2+2 \sqrt{-3}+\left(-1-\sqrt{-3}+ \sqrt{5}+\sqrt{-15}\right) x+4 x^2\right)$$ so the claim follows from the Brahmagupta identity.

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