Identity

$a - b$ divides $a^{n} - b^{n}$ for every $n$

Algebra • Expressions • Algebraic Identities

Formula

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

For every 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 all carrying a plus sign; in particular

a - b

divides

a^{n} - b^{n}

for every

n

, odd or even.

How to use it

1In a divisibility question, first force both terms to carry the same exponent — $5^{2 n} = 2 5^{n}$ and $2^{3 n} = 8^{n}$ — and then subtract the bases.
2Equivalently, modulo $a - b$ one has $b \equiv a$ , so $a^{n}$ and $b^{n}$ leave the same remainder and their difference is a multiple of $a - b$ .
3 $n = 2$ is the difference of squares and $n = 3$ the difference of cubes; every higher exponent behaves the same way, with no condition on $n$ .

Show that

8^{n} - 3^{n}

is a multiple of

5

for every positive integer

n

Step-by-step

1
$8 - 3 = 5$
Why: The difference of the bases is the factor the identity produces.
2
$8^{n} - 3^{n} = (8 - 3) (8^{n - 1} + 8^{n - 2} \cdot 3 + \dots + 3^{n - 1}) = 5 \times (an integer)$

Answer

8^{n} - 3^{n}

is divisible by

5

for every positive integer

n

Watch out

The companion result for sums is weaker:

a + b

divides

a^{n} + b^{n}

only when

n

is odd, and

a^{2} + b^{2}

has no such factor at all. Interchanging the two is the usual error. Nothing here says

a - b

is the largest factor —

a^{n} - b^{n}

is usually divisible by much more.

Builds on

Also known as

difference of like powers, a minus b divides a^n minus b^n, factorisation of a^n - b^n, difference of nth powers, difference of equal powers