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I have to proof that $X^n+1$ is divisible by $X+1$ if $n$ is odd ($n\in\mathbb N$). If $n$ is odd, that means that $n=2m+1$ ($m\in\mathbb N_0$). I was using full induction.

Base: If $m=0$ then $X^{2m+1}+1$ is $X^1+1$, which is divisible by $X+1.$

Assumption: $X^{2m+1}+1$ is divisible by $X+1$, true.

Step: I set $m$ to $m+1$, which means I get $X^{2(m+1)+1}+1$ which equals $X^{2m+3}+1$. And that's the point where I'm having trouble. I tried several ways to split it up, but I never got it in a way so that I could use the assumption for the entire equation, just for a few parts of it, at best.

I really don't know what to do at this point, maybe my approach is wrong, or it's something else. Whatever it is, I really would appreciate if someone could lead me onto the right path.

Bill Dubuque
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hearteyes
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2 Answers2

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We have the direct factorization $$ x^n + 1 =(x+1)(1 - x + x^2 - x^3 + \ldots + x^{n-1}) $$ which can be checked by expanding the right-hand side and noting the cascading cancellations. (We are using that $n$ is odd because the sign on the term with exponent $n-1$ needs to be positive.)

Arthur
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hunter
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  • Just for clarification (since the OP wanted to use induction): The "$\ldots$" in there is where you actually use induction to do the calculation. – Arthur Nov 12 '22 at 15:12
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 12 '22 at 16:41
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Use

$$x^{2m+3} + 1 = x^{2m+1} + 1 + (x + 1)(x^{2m+2} - x^{2m+1})$$

JMP
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