Prove that if m is odd, then m^n is odd for n consisting of all natural numbers?

Question

Prove that if m is odd, then m^n is odd for n consisting of all natural numbers?

Answer 1

If m is odd it can written as 2k+1 for some other number k.

(for example, 51 = 2 * 25 + 1)

Then (2k+1)^n when expanded looks like

(2k)^n + a bunch of terms a_i * (2k)^i (1)^(n-i) + ... + 1

All terms but the last have a factor 2k and therefore a factor 2

So their sum is even or 2j for some number j.

Therefore the total is 2j+1 which is again an odd number.

You could also do it by induction:

Let d be an odd number.

d^1 = d, which is odd.

d^k is odd by induction hypothesis

So, let d^k = 2j + 1 for some j

Multiply by d = 2m + 1 for some m

(2j + 1) (2m + 1) = 4 jm + 2j + 2m + 1

where the sum of the first three terms is even, 2p for some p,

and then we add the last 1, so we have 2p+1 which is odd.

Therefore d^k * d = d^(k+1) is odd, and since d^1 is odd,

all d^n are odd if d is odd.

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