Difference of Two Powers/General Commutative Ring

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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:

\(\displaystyle a^n - b^n\) \(=\) \(\displaystyle \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }\)


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:

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

$\blacksquare$