Difference of Two Powers

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.

Let $a, b \in R$.

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

Then:

When $R$ is one of the standard number systems, that is $\Z, \Q, \R$ and so on, this translates into:

For convenience of applicability, these results are sometimes separated into two cases for odd and even indices:

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

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

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

Taking the first term (where $j = 0$) out of the summation, we get:
 * $\displaystyle a \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} \circ b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$

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

Taking the first term (where $j = 0$) out of the summation:
 * $\displaystyle b \circ S_n = \sum_{j \mathop = 0}^{n - 1} a^j \circ b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} \circ b^j$

This is equal to:
 * $\displaystyle b^n + \sum_{j \mathop = 1}^{n - 1} a^j \circ b^{n - j}$

by Permutation of Indices of Summation.

So: