Algebraic factorizations

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For positive integer

n > 1

, we have

a n − b n  =  (a − b)     n ∑   i   = 1    a n  − i b i  − 1  =  (a − b) (a n  − 1 + a n  − 2 b + a n  − 3 b 2 + ⋯ + a b n  − 2 + b n  − 1).

For odd positive integer

2 n + 1, n > 1,

we have

a 2 n +1 + b 2 n +1  =  (a + b)   2 n +1 ∑   i   = 1    a 2 n +1 − i (−b) i  − 1  =  (a + b) (a 2 n − a 2 n  − 1 b + a 2 n  − 2 b 2 + ⋯ − a b 2 n  − 1 + b 2 n ).

For even positive integer

2 n

, we have

a 2 n + b 2 n  =  (a n − 2√  2 a n b n + b n ) (a n + 2√  2 a n b n + b n ).

Examples

Table of algebraic factorizations

n	a n  −  b n, a n + b n.
2	${a^{2}-b^{2}=(a-b)(a+b)},\,$ ${a^{2}+b^{2}=(a-{\sqrt {2ab}}+b)(a+{\sqrt {2ab}}+b)}.\,$
3	${a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})},\,$ ${a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})=(a-{\sqrt {3ab}}+b)(a+b)(a+{\sqrt {3ab}}+b)}.\,$
4	${a^{4}-b^{4}=(a^{2}-b^{2})(a^{2}+b^{2})=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3})=(a-b)(a+b)(a^{2}+b^{2})},\,$ ${a^{4}+b^{4}=(a^{2}-{\sqrt {2}}\,ab+b^{2})(a^{2}+{\sqrt {2}}\,ab+b^{2})}.\,$
5	${a^{5}-b^{5}=(a-b)(a^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4})},\,$ ${a^{5}+b^{5}=(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4})}.\,$
6	${a^{6}-b^{6}=(a^{3}-b^{3})(a^{3}+b^{3})=(a^{2}-b^{2})(a^{4}+a^{2}b^{2}+b^{4})=(a-b)(a+b)(a^{2}-ab+b^{2})(a^{2}+ab+b^{2})},\,$ ${a^{6}+b^{6}=(a^{2}+b^{2})(a^{4}-a^{2}b^{2}+b^{4})}.\,$
7	${a^{7}-b^{7}=(a-b)(a^{6}+a^{5}b+a^{4}b^{2}+a^{3}b^{3}+a^{2}b^{4}+ab^{5}+b^{6})},\,$ ${a^{7}+b^{7}=(a+b)(a^{6}-a^{5}b+a^{4}b^{2}-a^{3}b^{3}+a^{2}b^{4}-ab^{5}+b^{6})}.\,$
8	${a^{8}-b^{8}=(a^{4}-b^{4})(a^{4}+b^{4})=(a-b)(a+b)(a^{2}+b^{2})(a^{4}+b^{4})},\,$ ${a^{8}+b^{8}=(a^{4}-{\sqrt {2}}\,a^{2}b^{2}+b^{4})(a^{4}+{\sqrt {2}}\,a^{2}b^{2}+b^{4})}.\,$
9	${a^{9}-b^{9}=(a^{3}-b^{3})(a^{6}+a^{3}b^{3}+b^{6})=(a-b)(a^{2}+ab+b^{2})(a^{6}+a^{3}b^{3}+b^{6})},\,$ ${a^{9}+b^{9}=(a^{3}+b^{3})(a^{6}-a^{3}b^{3}+b^{6})=(a+b)(a^{2}-ab+b^{2})(a^{6}-a^{3}b^{3}+b^{6})}.\,$
10	${a^{10}-b^{10}=(a^{5}-b^{5})(a^{5}+b^{5})=(a-b)(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4})(a^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4})},\,$ ${a^{10}+b^{10}=(a^{2}+b^{2})(a^{8}-a^{6}b^{2}+a^{4}b^{4}-a^{2}b^{6}+b^{8})}.\,$
11	${a^{11}-b^{11}=(a-b)(a^{10}+a^{9}b+a^{8}b^{2}+a^{7}b^{3}+a^{6}b^{4}+a^{5}b^{5}+a^{4}b^{6}+a^{3}b^{7}+a^{2}b^{8}+ab^{9}+b^{10})},\,$ ${a^{11}+b^{11}=(a+b)(a^{10}-a^{9}b+a^{8}b^{2}-a^{7}b^{3}+a^{6}b^{4}-a^{5}b^{5}+a^{4}b^{6}-a^{3}b^{7}+a^{2}b^{8}-ab^{9}+b^{10})}.\,$
12	${\begin{aligned}{a^{12}-b^{12}}&={(a^{6}-b^{6})(a^{6}+b^{6})=(a^{3}-b^{3})(a^{3}+b^{3})(a^{6}+b^{6})=(a^{3}-b^{3})(a^{3}+b^{3})(a^{2}+b^{2})(a^{4}-a^{2}b^{2}+b^{4})}\\&={(a-b)(a+b)(a^{2}+b^{2})(a^{2}-ab+b^{2})(a^{2}+ab+b^{2})(a^{4}-a^{2}b^{2}+b^{4})},\end{aligned}}$ ${a^{12}+b^{12}=(a^{4}+b^{4})(a^{8}-a^{4}b^{4}+b^{8})}.\,$

Algebraic factorization of integers

629 = 625 + 4 = 25 2 + 2 2 = (25  − 

2√  2  ×  25  ×  2

+ 2) (25 +

2√  2  ×  25  ×  2

+ 2) = 17  ×  37.

117 = 125  −  8 = 5 3  −  2 3 = (5  −  2) (5 2  +  5  ×  2 + 2 2 ) = 3  ×  39.

133 = 125  +  8 = 5 3  +  2 3 = (5  +  2) (5 2  −  5  ×  2 + 2 2 ) = 7  ×  19.

629 = 625 + 4 = 5 4 + 2 2 = 5 4 + (

2√  2

 ) 4 = (5 2  − 

2√  2

 ×  5  × 

2√  2

+ (

2√  2

 ) 2 ) (5 2 +

2√  2

 ×  5  × 

2√  2

+ (

2√  2

 ) 2 ) = (25  −  10 + 2) (25 + 10 + 2) = 17  ×  37.

390629 = 390625 + 4 = 5 8 + 2 2 = 5 8 + ( 

4√  2

 ) 8 = (5 4  − 

2√  2

 ×  5 2  ×  ( 

4√  2

 ) 2 + ( 

4√  2

 ) 4 ) (5 4 +

2√  2

 ×  5 2  ×  ( 

4√  2

 ) 2 + ( 

4√  2

 ) 4 ) = (625  −  50 + 2) (625 + 50 + 2) = 577  ×  677.

See also

• Aurifeuillean factorizations

• Repunits (base

  b

  ) and repdigits (base

  b

  ) (when the number of digits is composite, we have algebraic factorizations)