Identity

$a + b$ divides $a^{n} + b^{n}$ for odd $n$

Algebra • Expressions • Algebraic Identities

Formula

a^{n} + b^{n} = (a + b) (a^{n - 1} - a^{n - 2} b + \dots + b^{n - 1})

For every odd positive integer

n

a^{n} + b^{n} = (a + b) (a^{n - 1} - a^{n - 2} b + a^{n - 3} b^{2} - \dots + b^{n - 1})

, the second factor having

n

terms with alternating signs; in particular

a + b

divides

a^{n} + b^{n}

whenever

n

is odd.

How to use it

1In a remainder question, look for bases that add to the modulus: every such pair contributes a multiple of it and drops out of the sum.
2Equivalently, work with negative residues - modulo $a + b$ one has $b \equiv - a$ , and an odd power preserves the sign.
3 $n = 3$ is the familiar sum of cubes; $n = 5$ , $n = 7$ and every other odd exponent behave in exactly the same way.

What is the remainder when

4^{7} + 3^{7}

is divided by

7

Step-by-step

1
$4 + 3 = 7, and the exponent 7 is odd$
Why: An odd exponent is the condition for the factorisation.
2
$4^{7} + 3^{7} = (4 + 3) (4^{6} - 4^{5} \cdot 3 + \dots + 3^{6})$
Why: The sum of the bases comes out as a factor.
3
$7 ∣ (4^{7} + 3^{7}) \Rightarrow remainder = 0$

Answer

0

Watch out

Odd

n

only:

a^{2} + b^{2}

has no such factor at all. The companion result is different -

a - b

divides

a^{n} - b^{n}

for EVERY

n

, odd or even - and interchanging the two is the usual error.

Builds on

Also known as

sum of odd powers, a plus b divides a^n plus b^n, factorisation of a^5 + b^5, odd power sum factorisation