# Talk:Factors of Sum of Two Even Powers

In my copy of 1968: Murray R. Spiegel: *Mathematical Handbook of Formulas and Tables* ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.23$ the final product has a $\paren {2 n - 1} \pi$ rather than $\paren {2 n + 1} \pi$ factor. As such, we may wish to modify the theorem statement to more accurately reflect what is written in the source. Both statements are true and the middle term can be positive or negative, and it may be beneficial to prove the "harder" version as a demonstration of the symmetries available.

I'm not convinced that the Factorisation of $z^n - a$ is the best starting point as it's a bit of a leap to use the complex $2 n$th roots of $-1$ given it's reliance upon the nth roots of $+1$ rather than $-1$. Within Factorisation of $z^n - a$ there are a number of references to $\theta$ which aren't defined - they're all 0 as a is a real number. I'm happy to correct that anomaly and to add another page for Factorisation of $z^n + a$.

IMHO the signs of the $\alpha$s do need to be reversed - the corrected proof will probably rely on the complex $2 n$th roots of $-1$, so we'd need to use minus instead of plus when we multiply the chosen pair of factors.

By pairing-off the factors from opposite ends of the list, e.g. $k$ with $\paren {2 n - k}$ I expect that the middle term in the product will be $- 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n}$. It may be possible to find a more elegant pairing, e.g. by leveraging the symmetry that appears due to taking an even-powered root, i.e. any root in one quadrant is also reflected in the other three quadrants of the Argand diagram. Given that we still need to fix the $\paren {2 n - 1} \pi$ vs. $\paren {2 n + 1} \pi$ numerator and fix the sign of the middle term, both of these can be easily achieved by relabelling, e.g. setting $k = n + j - 1$ means $2k + 1 = 2n + 2j - 2 + 1 = 2n + 2j - 1$, without resorting to a more elegant set of pairings. The additional $2n$ in the numerator shifts the argument by $2 n \pi / 2n = \pi$ and $cos(x + \pi) = -cos(x)$ gives the required change of sign for the middle term.

I'm happy to make all the above changes and additions - this will be my first edit and I would like to make sure that there aren't any serious objections before I start, e.g. changing an existing proof in Factorisation of $z^n - a$ to remove/elucidate on the unexplained $\theta$s.

- Feel free to go ahead and fix what needs fixing. I lack the ability to apply sufficient concentration nowadays (getting old) so had trouble with it. --prime mover (talk) 18:42, 26 December 2019 (EST)