>>191
>複素数の範囲で因数分解

ふむ、下記のクンマー理論 背理法か
フェルマーの最終定理 a^n+b^n=c^n
これが複素数の範囲で因数分解できて
Cyclotomic polynomial 円分多項式の理論が使える
これぞ、クンマー・デデキント 理想数・イデアルぞ！（＾＾
素因数分解の一意性と類似が成り立ち、背理法が使える！！！
となったらしい
詳しくは下記など

(参考)
https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
Fermat's Last Theorem
Proofs for specific exponents
Early modern breakthroughs
Ernst Kummer and the theory of ideals
(It is often stated that Kummer was led to his "ideal complex numbers" by his interest in Fermat's Last Theorem; there is even a story often told that Kummer, like Lamé, believed he had proven Fermat's Last Theorem until Lejeune Dirichlet told him his argument relied on unique factorization; but the story was first told by Kurt Hensel in 1910 and the evidence indicates it likely derives from a confusion by one of Hensel's sources. Harold Edwards said the belief that Kummer was mainly interested in Fermat's Last Theorem "is surely mistaken".[144] See the history of ideal numbers.)

https://math.stackexchange.com/questions/3684655/factorization-of-anbn
stackexchange
asked May 21, 2020 user730322
Factorization of a^n+b^n
I am trying to figure out a way to factor a^n+b^n,
but all I found is odd cases where a^n+b^n=(a+b)(a^n−1 −a^n−2 b +...−ab^n−2 + b^n−1).

1 Answer answered May 21, 2020 Angina Seng
Let's look at X^n+1 instead. Then
X^n+1=(X^2n −1)/(X^n−1).
Over the rationals, the irreducible factorisation of X^n−1 is
X^n −1=∏d∣n Φd(X)
where Φd is the d-th cyclotomic polynomial. Therefore
X^n +1=(∏d∣n Φ2d(X))/(∏d∣n Φd(X))=∏d∣2n,d∤n Φd(X).
Now homogenise:
an+bn=∏d∣2n,d∤n Φd(a,b)
where
Φd(X,Y)=Y^degΦd Φd(X/Y).

https://en.wikipedia.org/wiki/Cyclotomic_polynomial
Cyclotomic polynomial

つづく