Given a commutative ring $(R,+,*)$ I would like to understand if the equality $$ \tag{1}\label{1}x^n-y^n =(x-y)\sum_{i=0}^{n-1} x^i y^{n-1-i} $$ holds for any $x,y\in R$. In fact, here I found a proof of the above identity where $x$ and $y$ are reals but actually the proof is by induction using only the usual properties of any commutative ring so that I believe \eqref{1} holds for any commutative ring: however, into the comments the professor Brandemburg seems state that induction does not work so that I thought to put here an answer where I ask if \eqref{1} holds for commutative ring and so if it is possibile to prove it with induction. So clould someone help me, please?
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user26857
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Antonio Maria Di Mauro
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(1) will be true. In the post that you have linked, Martin Brandenburg mentions that induction is overkill meaning it involves a lot more steps than is actually needed to prove the result. A proof using induction will still work here. – Kendall Jun 30 '23 at 14:54
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No, Brandemburg does not state induction does not work. – coiso Jun 30 '23 at 14:55
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@elemelons The issue is that Brandemburg stated what to follow: «Also, in general induction will only work if you already know one side of some equation. What happens when you really want to find some expression?» and this confused me since $x$ and $y$ are not kown. – Antonio Maria Di Mauro Jun 30 '23 at 15:07
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@AntonioMariaDiMauro Brandemburg is talking about how you find formulas to begin with. You can prove the geometric sum formula, but where did the formula come from? Typically, teachers just hand students formulas like this. Which, depending on the situation, may not be the best pedagogy because it does not explain the formula or prepare students for trying to discover things on their own. – coiso Jun 30 '23 at 15:20
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@elemelons Okay, so you are stating that Brandemburg comment is only an educational comment, right? – Antonio Maria Di Mauro Jun 30 '23 at 15:29
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1@Kendall Induction on $n$ (or equivalent) is necessary for a rigorous proof. The remark that it is "overkill" is misleading. Note that the induction might be hidden somewhere down a chain of inferences, e.g. invoklng another theorem with inductive proof. Possibly that's what M.B. meant. – Bill Dubuque Jun 30 '23 at 15:31
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Yes. (And I agree with Bill's comment about induction being necessary for rigor.) – coiso Jun 30 '23 at 15:33