Difference of Two Powers

Theorem

Let $\left({R, +, \circ}\right)$ be a commutative ring whose zero is $0_R$.

Let $a, b \in R$.

Let $n \in \N$ such that $n \ge 2$.

Then:

$\displaystyle a^n - b^n = \left({a - b}\right) \circ \left({a^{n-1} + a^{n-2} \circ b + a^{n-3} \circ b^2 + \ldots + a \circ b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \circ \sum_{j=0}^{n-1} a^{n-j-1} \circ b^j$

When $R$ is one of the standard sets of numbers, i.e. $\Z, \Q, \R$ etc., then this translates into:

$\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \ldots + a b^{n-2} + b^{n-1}}\right) = \left({a - b}\right) \sum_{j=0}^{n-1} a^{n-j-1} b^j$

Proof

Let $\displaystyle S_n = \sum_{j=0}^{n-1} a^{n-j-1} \circ b^j$.

This can also be written $\displaystyle S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j-1}$.

Consider $\displaystyle a \circ S_n = \sum_{j=0}^{n-1} a^{n-j} \circ b^j$.

Taking the first term (where $j = 0$) out of the summation, we get:

$\displaystyle a \circ S_n = \sum_{j=0}^{n-1} a^{n-j} \circ b^j = a^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j$

Similarly, consider $\displaystyle b \circ S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j}$.

Taking the first term (where $j = 0$) out of the summation:

$\displaystyle b \circ S_n = \sum_{j=0}^{n-1} a^j \circ b^{n-j} = b^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j$

This is equal to $\displaystyle b^n + \sum_{j=1}^{n-1} a^j \circ b^{n-j}$ by permutation of indices.

So:

	\(\displaystyle \)	\(\displaystyle \left({a - b}\right) \circ S_n\)	\(=\)	\(\displaystyle a \circ \sum_{j=0}^{n-1} a^{n-j-1} \circ b^j - b \circ \sum_{j=0}^{n-1} a^j \circ b^{n-j-1}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{j=0}^{n-1} a^{n-j} \circ b^j - \sum_{j=0}^{n-1} a^j \circ b^{n-j}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^n + \sum_{j=1}^{n-1} a^{n-j} \circ b^j - \sum_{j=1}^{n-1} a^{n-j} \circ b^j - b^n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^n - b^n\)	\(\displaystyle \)

$\blacksquare$

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $3$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 3.11 \ (2)$