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The question is : Prove by mathematical induction that :

$$ a^{2n-1} + b^{2n+1} $$

is divisible by $$a+b$$

I've done a lot of stuff but can't put them down in tex properly. Thanks.

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    $\begingroup$ Then at least explain what your work is. Your question is too vague. $\endgroup$ – Andrea Orta Oct 1 '12 at 14:41
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    $\begingroup$ How is it "too vague"? $\endgroup$ – Mob Oct 1 '12 at 14:41
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    $\begingroup$ If you have "done a lot of stuff", then probably you don't need a generic hint, but something more specific. If you neither show nor explain your first results, we can't distinguish your question from the one of somebody who (for example) didn't understand induction at all and couldn't even start a solution. Then our advice won't be so effective as it could be. $\endgroup$ – Andrea Orta Oct 1 '12 at 14:49
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    $\begingroup$ Is that supposed to be $a^{2n+1}+b^{2n+1}$ instead of $a^{2n-1}+b^{2n+1}$, perhaps? $\endgroup$ – Cameron Buie Oct 1 '12 at 14:56
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    $\begingroup$ @Mob The exponents must be equal else it is false, e.g. for $\rm\:n=1,\ a=2=b,\ a^{2n-1}\!+b^{2n+1} = 10\:$ is not divisible by $\rm\:a+b = 4.\ \ $ $\endgroup$ – Bill Dubuque Oct 1 '12 at 15:21
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Write it as

$$a^{2n+1}+b^{2n+1}=(a^2+b^2)(a^{2n-1}+b^{2n-1})-a^2b^2(a^{2n-3}+b^{2n-3})$$ and then use induction on $n$.

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  • $\begingroup$ Thanks please, I don't understand. $\endgroup$ – Mob Oct 1 '12 at 14:44
  • $\begingroup$ +1: Mob, you'll actually be using a strong induction on $n$. Show that your proposition holds for $n=1,n=2$, and then you'll use the above expression to show that if $a+b$ divides $$a^{2n-3}+b^{2n-3}=a^{2(n-2)+1}+b^{2(n-2)+1}$$ and $$a^{2n-1}+b^{2n-1}=a^{2(n-1)+1}+b^{2(n-1)+1},$$ then it divides $a^{2n+1}+b^{2n+1}$. $\endgroup$ – Cameron Buie Oct 1 '12 at 14:57
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By induction $\rm\:(-b)^{2n-1}\! = - b^{2n-1}$ so $\rm\:mod\ a\!+\!b\!:\ a\equiv-b\:\Rightarrow\:a^{2n-1}\!\equiv(-b)^{2n-1}\!\equiv {-}b^{2n-1}$

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A mild variant of the answer by dmm: $$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}+b^{2n})-ab(a^{2n-1}+b^{2n-1}).$$

Then directly from $a+b$ divides $a^{2k-1}+b^{2k-1}$ we can conclude that $a+b$ divides $a^{2k+1}+b^{2k+1}$. This may feel more comfortably familiar.

Added: In answer to a comment, the OP has indicated that $a^{2n-1}+b^{2n+1}$, which looked like an obvious typo, is not. If so, the conjectured result is false. Let $a=2$, $b=3$, $n=1$.

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Hint:

$$\forall\,n\in\Bbb N\,\,\,,\,\,a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+....-ab^{2n-1}+b^{2n})$$

Formally, the above still requires a little proof by induction.

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