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Prove that for all integers $a$ and $b$ and all $n \in \mathbb N$,

$$(a-b) \mid (a^{n}-b^{n})$$

My attempt:

Take arbitrary $a,b \in \mathbb Z$.

By induction.

Base case: $n = 0$

$a^0 - b^0 = 1 - 1 = 0$ And $(a-b) \mid 0$

Suppose proposition holds for $n$, i.e

$$(a-b) \mid (a^{n}-b^{n})$$

Consider

$$\begin{align} a^{n+1} - b^{n+1} & = aa^{n} - bb^{n}\\ & = aa^{n} - bb^{n} + ba^{n} - ba^{n} \\ & = aa^{n} -ba^{n}+ba^n - bb^n \\ & = a^n(a - b) + b(a^{n} - b^{n}) \end{align}$$

Since $(a-b )\mid a^{n} - b^{n}$, we can write

$$a^{n} - b^{n} = k(a-b), k \in \mathbb Z$$

Substituting gives

$$a^n(a - b) + bk(a-b) = (a-b)(a^{n} + bk)$$ Hence $a^{n+1} - b^{n+1}$ is divisible by $(a-b)$. $\Box$

Is it correct?

Stokolos Ilya
  • 3,371

1 Answers1

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Your proof is correct.

It is however easier to note that $\frac{a^{n+1} - b^{n+1}}{a-b}$ is the sum of the geometric progression $a^{n}+a^{n-1}b+ ... +b^{n}.$