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Suppose $a$ is some positive odd integer. Prove that $\forall n \in \mathbb{N}$, $a^n$ is odd.

I assume we define $a$ to be an odd positive, so $a = 2k+1 > 0$. Then we use induction or something of the sorts to show that $(2k+1+1)^n$ is also odd. How do we go about this, without knowing $n$?

Would we split this up into cases where $n$ is odd and $n$ is even?

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closed as off-topic by Namaste, Leucippus, Antonios-Alexandros Robotis, Shailesh, José Carlos Santos Aug 11 '17 at 6:22

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  • $\begingroup$ Binomial formula. $\endgroup$ – Cauchy Jul 26 '17 at 14:33
  • $\begingroup$ Can you elaborate on that? $\endgroup$ – 1011011010010100011 Jul 26 '17 at 14:34
  • $\begingroup$ Uhm, $(2k+1+1)^n$ is even. $\endgroup$ – steven gregory Jul 26 '17 at 14:41
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Use induction and the fact that the product of two odd integers is odd.

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$$(2k+1)^n = \sum_{t = 0}^{n} {n \choose t} (2k)^t = 1 + \sum_{t=1}^m {n \choose t} 2^t k^t$$

the rightmost term is obviously even, so the result is odd.

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  • $\begingroup$ But this is proving it for all $a$. I need to prove it for all $n$. $\endgroup$ – 1011011010010100011 Jul 26 '17 at 14:38
  • $\begingroup$ @sgerbhctim this proves it for all $a$ and for all $n$. $\endgroup$ – Cauchy Jul 26 '17 at 14:39
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You said it yourself - proof by induction. Prove that if it's true for $n$, it is also true for $n+1$, then prove it's true for $n=1$.

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odd * odd = odd so any product of odds is odd: we know it's true for a product of $n=2\,$ odds, and a product of $n\!+\!1$ odds is an odd times a product of $n$ odds, so equals odd * odd = odd by induction.

Remark $ $ Exactly the same inductive proof shows that if a set $S$ is closed under multiplication then $S$ is also closed under $n$-ary products, i.e. under products of any length. Note that the proof uses only associativity of multiplication, so it works for any associative operator, i.e. in any semigroup or monoid.

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