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I know that $a-b$ can divide $a^n-b^n$ and I have already seen other proofs in a similar post here. But, I want to know if the following way of proving it is correct or not.

Say we have a G.P. : $1, \frac{b}{a},\frac{b^2}{a^2},\frac{b^3}{a^3} , \cdot \cdot \cdot , \frac{b^{n-1}}{a^{n-1}}$

Where, $a$ and $b$ are integers and $a\neq0$

If we add all $n$ terms, we get: $1+\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3} + \cdot \cdot \cdot + \frac{b^{n-1}}{a^{n-1}} = \frac{1\cdot\left ( 1 - \left ( \frac{b}{a} \right )^n \right )}{ \left ( 1 - \left ( \frac{b}{a} \right ) \right )}$

This can be rewritten as:$1+\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3} + \cdot \cdot \cdot + \frac{b^{n-1}}{a^{n-1}} = \frac{a^{1-n}(a^n - b^n)}{(a - b)}$

And then after a few operations it becomes: $ a^{n-1} + a^{n-2}b + \cdot \cdot \cdot + b^{n-1} = \frac{(a^n - b^n)}{(a - b)}$

And because $a, b \in \mathbb{I}$, the L.H.S. is also an integer, which tells us that $a-b$ can divide $a^n-b^n$ regardless if $n$ is odd or even.

Is this way of proving it correct? Because, I find it rather simpler than other methods discussed here: Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd

dinoweb69
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  • https://math.stackexchange.com/questions/tagged/solution-verification: For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so. – Anne Bauval Nov 04 '24 at 14:29
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    I don't find it simpler than https://math.stackexchange.com/a/17796 – Anne Bauval Nov 04 '24 at 14:32
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    Simpler: The equation $x^n-a^n=0$ has the obvious solution $x=a$, hence the polynomial $x^n-a^n$ is a multiple of $x-a$. (You find the quotient by long division.) –  Nov 04 '24 at 14:40
  • Not simpler either (than the linked answer). – Anne Bauval Nov 04 '24 at 15:33
  • I am trying to prove it here. I am not using a trick as it is mentioned in the linked answer. I think this can itself serve as a proof of the telescopic series mentioned. It is really reverse engineering of the telescopic series. – dinoweb69 Nov 04 '24 at 16:24
  • @dinoweb69: the formula for the geometric series is usually obtained by telescoping. –  Nov 04 '24 at 17:39
  • Yes it is completely correct. I also used to prove results like this with this GP method. – Blue Cat Blues Nov 04 '24 at 12:39

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This proof involves two hidden inductions (basically, anywhere a "..." appears!); if your students/readers don't know how to do proof by induction, then I'd say it's not great.

The last step is also needlessly complicated. You could instead multiply through and say $$ (a-b)(a^{n-1} + a^{n-2}b + \cdot \cdot \cdot + b^{n-1}) = (a^n - b^n) $$ which provides an explicit factorization of $a^n - b^n$, with $a-b$ as one of the factors, showing it's definitely a divisor.

John Hughes
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  • I'm a student(high-school) myself. I know the induction method, but I wanted to find alternatives. As for the last step, I thought it best to state everything explicitly. – dinoweb69 Nov 04 '24 at 13:20
  • I agree. And it's the implicit inductions here that make your proof not really better than alternatives. – John Hughes Nov 04 '24 at 15:59