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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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    @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.\ \ $2012-10-01

4 Answers 4

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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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    +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 $a^{2n+1}+b^{2n+1}$.2012-10-01
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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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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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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.