• Numbers of the form ${\displaystyle a^{4}+4b^{4}}$  have the following factorization (Sophie Germain's identity):
  $${\displaystyle a^{4}+4b^{4}=(a^{2}-2ab+2b^{2})\cdot (a^{2}+2ab+2b^{2}).}$$

   
  Setting ${\displaystyle a=1}$  and ${\displaystyle b=2^{k}}$ , one obtains the following aurifeuillean factorization of ${\displaystyle \Phi _{4}(2^{2k+1})=2^{4k+2}+1}$ , where ${\displaystyle \Phi _{4}(x)=x^{2}+1}$  is the fourth cyclotomic polynomial:^[2]
  $${\displaystyle 2^{4k+2}+1=(2^{2k+1}-2^{k+1}+1)\cdot (2^{2k+1}+2^{k+1}+1).}$$

   
• Numbers of the form ${\displaystyle a^{6}+27b^{6}}$  have the following factorization, where the first factor (${\displaystyle a^{2}+3b^{2}}$ ) is the algebraic factorization of sum of two cubes:
  $${\displaystyle a^{6}+27b^{6}=(a^{2}+3b^{2})\cdot (a^{2}-3ab+3b^{2})\cdot (a^{2}+3ab+3b^{2}).}$$

   
  Setting ${\displaystyle a=1}$  and ${\displaystyle b=3^{k}}$ , one obtains the following factorization of ${\displaystyle 3^{6k+3}+1}$ :^[2]
  $${\displaystyle 3^{6k+3}+1=(3^{2k+1}+1)\cdot (3^{2k+1}-3^{k+1}+1)\cdot (3^{2k+1}+3^{k+1}+1).}$$

   
  Here, the first of the three terms in the factorization is ${\displaystyle \Phi _{2}(3^{2k+1})}$  and the remaining two terms provide an aurifeuillean factorization of ${\displaystyle \Phi _{6}(3^{2k+1})}$ , where ${\displaystyle \Phi _{6}(x)=x^{2}-x+1}$ .
• Numbers of the form ${\displaystyle b^{n}-1}$  or their factors ${\displaystyle \Phi _{n}(b)}$ , 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:
  • ${\displaystyle t\equiv 1{\pmod {4}}}$  and ${\displaystyle n\equiv t{\pmod {2t}}}$ 
  • ${\displaystyle t\equiv 2,3{\pmod {4}}}$  and ${\displaystyle n\equiv 2t{\pmod {4t}}}$ 

Thus, when ${\displaystyle b=s^{2}\cdot t}$  with square-free ${\displaystyle t}$ , and ${\displaystyle n}$  is congruent to ${\displaystyle t}$  modulo ${\displaystyle 2t}$ , 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), aurifeuillean factorization may be used, which gives a product of two or three numbers. The following equations give aurifeuillean factors for the Cunningham project bases as a product of F, L and M:^[3]

If we let L = C − D, M = C + D, the aurifeuillean factorizations for bⁿ ± 1 of the form F * (C − D) * (C + D) = F * L * M 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 and up to 998, see ^[4][5][6])

b	Number	(C − D) * (C + D) = L * M	F	C	D
2	24k + 2 + 1	${\displaystyle \Phi _{4}(2^{2k+1})}$	1	22k + 1 + 1	2k + 1
3	36k + 3 + 1	${\displaystyle \Phi _{6}(3^{2k+1})}$	32k + 1 + 1	32k + 1 + 1	3k + 1
5	510k + 5 - 1	${\displaystyle \Phi _{5}(5^{2k+1})}$	52k + 1 - 1	54k + 2 + 3(52k + 1) + 1	53k + 2 + 5k + 1
6	612k + 6 + 1	${\displaystyle \Phi _{12}(6^{2k+1})}$	64k + 2 + 1	64k + 2 + 3(62k + 1) + 1	63k + 2 + 6k + 1
7	714k + 7 + 1	${\displaystyle \Phi _{14}(7^{2k+1})}$	72k + 1 + 1	76k + 3 + 3(74k + 2) + 3(72k + 1) + 1	75k + 3 + 73k + 2 + 7k + 1
10	1020k + 10 + 1	${\displaystyle \Phi _{20}(10^{2k+1})}$	104k + 2 + 1	108k + 4 + 5(106k + 3) + 7(104k + 2) + 5(102k + 1) + 1	107k + 4 + 2(105k + 3) + 2(103k + 2) + 10k + 1
11	1122k + 11 + 1	${\displaystyle \Phi _{22}(11^{2k+1})}$	112k + 1 + 1	1110k + 5 + 5(118k + 4) - 116k + 3 - 114k + 2 + 5(112k + 1) + 1	119k + 5 + 117k + 4 - 115k + 3 + 113k + 2 + 11k + 1
12	126k + 3 + 1	${\displaystyle \Phi _{6}(12^{2k+1})}$	122k + 1 + 1	122k + 1 + 1	6(12k)
13	1326k + 13 - 1	${\displaystyle \Phi _{13}(13^{2k+1})}$	132k + 1 - 1	1312k + 6 + 7(1310k + 5) + 15(138k + 4) + 19(136k + 3) + 15(134k + 2) + 7(132k + 1) + 1	1311k + 6 + 3(139k + 5) + 5(137k + 4) + 5(135k + 3) + 3(133k + 2) + 13k + 1
14	1428k + 14 + 1	${\displaystyle \Phi _{28}(14^{2k+1})}$	144k + 2 + 1	1412k + 6 + 7(1410k + 5) + 3(148k + 4) - 7(146k + 3) + 3(144k + 2) + 7(142k + 1) + 1	1411k + 6 + 2(149k + 5) - 147k + 4 - 145k + 3 + 2(143k + 2) + 14k + 1
15	1530k + 15 + 1	${\displaystyle \Phi _{30}(15^{2k+1})}$	1514k + 7 - 1512k + 6 + 1510k + 5 + 154k + 2 - 152k + 1 + 1	158k + 4 + 8(156k + 3) + 13(154k + 2) + 8(152k + 1) + 1	157k + 4 + 3(155k + 3) + 3(153k + 2) + 15k + 1
17	1734k + 17 - 1	${\displaystyle \Phi _{17}(17^{2k+1})}$	172k + 1 - 1	1716k + 8 + 9(1714k + 7) + 11(1712k + 6) - 5(1710k + 5) - 15(178k + 4) - 5(176k + 3) + 11(174k + 2) + 9(172k + 1) + 1	1715k + 8 + 3(1713k + 7) + 1711k + 6 - 3(179k + 5) - 3(177k + 4) + 175k + 3 + 3(173k + 2) + 17k + 1
18	184k + 2 + 1	${\displaystyle \Phi _{4}(18^{2k+1})}$	1	182k + 1 + 1	6(18k)
19	1938k + 19 + 1	${\displaystyle \Phi _{38}(19^{2k+1})}$	192k + 1 + 1	1918k + 9 + 9(1916k + 8) + 17(1914k + 7) + 27(1912k + 6) + 31(1910k + 5) + 31(198k + 4) + 27(196k + 3) + 17(194k + 2) + 9(192k + 1) + 1	1917k + 9 + 3(1915k + 8) + 5(1913k + 7) + 7(1911k + 6) + 7(199k + 5) + 7(197k + 4) + 5(195k + 3) + 3(193k + 2) + 19k + 1
20	2010k + 5 - 1	${\displaystyle \Phi _{5}(20^{2k+1})}$	202k + 1 - 1	204k + 2 + 3(202k + 1) + 1	10(203k + 1) + 10(20k)
21	2142k + 21 - 1	${\displaystyle \Phi _{21}(21^{2k+1})}$	2118k + 9 + 2116k + 8 + 2114k + 7 - 214k + 2 - 212k + 1 - 1	2112k + 6 + 10(2110k + 5) + 13(218k + 4) + 7(216k + 3) + 13(214k + 2) + 10(212k + 1) + 1	2111k + 6 + 3(219k + 5) + 2(217k + 4) + 2(215k + 3) + 3(213k + 2) + 21k + 1
22	2244k + 22 + 1	${\displaystyle \Phi _{44}(22^{2k+1})}$	224k + 2 + 1	2220k + 10 + 11(2218k + 9) + 27(2216k + 8) + 33(2214k + 7) + 21(2212k + 6) + 11(2210k + 5) + 21(228k + 4) + 33(226k + 3) + 27(224k + 2) + 11(222k + 1) + 1	2219k + 10 + 4(2217k + 9) + 7(2215k + 8) + 6(2213k + 7) + 3(2211k + 6) + 3(229k + 5) + 6(227k + 4) + 7(225k + 3) + 4(223k + 2) + 22k + 1
23	2346k + 23 + 1	${\displaystyle \Phi _{46}(23^{2k+1})}$	232k + 1 + 1	2322k + 11 + 11(2320k + 10) + 9(2318k + 9) - 19(2316k + 8) - 15(2314k + 7) + 25(2312k + 6) + 25(2310k + 5) - 15(238k + 4) - 19(236k + 3) + 9(234k + 2) + 11(232k + 1) + 1	2321k + 11 + 3(2319k + 10) - 2317k + 9 - 5(2315k + 8) + 2313k + 7 + 7(2311k + 6) + 239k + 5 - 5(237k + 4) - 235k + 3 + 3(233k + 2) + 23k + 1
24	2412k + 6 + 1	${\displaystyle \Phi _{12}(24^{2k+1})}$	244k + 2 + 1	244k + 2 + 3(242k + 1) + 1	12(243k + 1) + 12(24k)

• Lucas numbers ${\displaystyle L_{10k+5}}$  have the following aurifeuillean factorization:^[7]

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

where ${\displaystyle L_{n}}$  is the ${\displaystyle n}$ th Lucas number, and ${\displaystyle F_{n}}$  is the ${\displaystyle n}$ th Fibonacci number.

In 1869, before the discovery of aurifeuillean factorizations, Landry [fr; es; de], through a tremendous manual effort,^[8][9] obtained the following factorization into primes:

${\displaystyle 2^{58}+1=5\cdot 107367629\cdot 536903681.}$ 

Three years later, in 1871, Aurifeuille discovered the nature of this factorization; the number ${\displaystyle 2^{4k+2}+1}$  for ${\displaystyle k=14}$ , with the formula from the previous section, factors as:^[2][8]

${\displaystyle 2^{58}+1=(2^{29}-2^{15}+1)(2^{29}+2^{15}+1)=536838145\cdot 536903681.}$ 

Of course, Landry's full factorization follows from this (taking out the obvious factor of 5). The general form of the factorization was later discovered by Lucas.^[2]

536903681 is an example of a Gaussian Mersenne norm.^[9]

• Aurifeuillean Factorisation, Colin Barker
• Aurifeuillean Factorizations, Gérard P. Michon
• The Search for Aurifeuillean-Like Factorizations
• Online factor collection
• A Note on Aurifeuillean Factorizations
• Aurifeuillean Factorisation