Aurifeuillian factor

An Aurifeuillian factor was named after the French mathematician Léon-François-Antoine Aurifeuille.

• Numbers of the form ${\displaystyle 2^{4k+2}+1}$ have the following Aurifeuillian factorization: Mathworld

${\displaystyle 2^{4k+2}+1=(2^{2k+1}-2^{k+1}+1)\cdot (2^{2k+1}+2^{k+1}+1)}$

• Numbers of the form ${\displaystyle b^n-1}$ or ${\displaystyle {\mathrm{\Phi }}_n(b)}$, where ${\displaystyle b=s^2\cdot t}$ with square-free ${\displaystyle t}$, have aurifeuillean factorization if and only if one of the following conditions holds:
  • ${\displaystyle t\equiv 1(\mathrm{mod}4)}$ and ${\displaystyle n\equiv t(\mathrm{mod}2t)}$
  • ${\displaystyle t\equiv 2,3(\mathrm{mod}4)}$ and ${\displaystyle n\equiv 2t(\mathrm{mod}4t)}$

Thus, when ${\displaystyle b=s^2\cdot t}$ with square-free ${\displaystyle t}$, and ${\displaystyle n}$ is congruent to ${\displaystyle t}$ mod ${\displaystyle 2t}$, then if ${\displaystyle t}$ is congruent to 1 mod 4, ${\displaystyle b^n-1}$ have aurifeuillean factorization, otherwise, ${\displaystyle b^n+1}$ have aurifeuillean factorization.

When the number is of a particular form (the exact expression varies with the base), Aurifeuillian factorization may be used, which gives a product of two or three numbers. The following equations give Aurifeuillian factors for the Cunningham project bases as a product of F, L and M: Main Cunningham Tables. At the end of tables 2LM, 3+, 5-, 6+, 7+, 10+, 11+ and 12+ are formulae detailing the Aurifeuillian factorisations.</ref>

If we let L = C - D, M = C + D, the Aurifeuillian factorizations for bⁿ ± 1 with the bases 2 ≥ b ≥ 24 (perfect powers excluded, since a power of bⁿ is also a power of b) are: (for the coefficients of the polynomials for all square-free bases up to 199, see Coefficients of Lucas C,D polynomials for all square-free bases up to 199)

(Number = F * (C - D) * (C + D) = F * L * M)

(See List of Aurifeuillean factorization for more information (square-free bases up to 199))

• Numbers of the form ${\displaystyle a^4+4b^4}$ have the following aurifeuillean factorization:

${\displaystyle a^4+4b^4=(a^2-2ab+2b^2)\cdot (a^2+2ab+2b^2)}$

• Lucas numbers ${\displaystyle L_{10k+5}}$ have the following aurifeuillean factorization: Lucas Aurifeuilliean primitive part

${\displaystyle L_{10k+5}=L_{2k+1}\cdot (5{F_{2k+1}}^2-5F_{2k+1}+1)\cdot (5{F_{2k+1}}^2+5F_{2k+1}+1)}$

where ${\displaystyle L_n}$ is the ${\displaystyle n}$th Lucas number, ${\displaystyle F_n}$ is the ${\displaystyle n}$th Fibonacci number.

External links

• Aurifeuillean factorization