I want to prove the following statements:

An odd positive integer raised to a positive power results in an odd number.


An even positive integer raised to a positive power results in an even number.

However, I am not sure how to do this mathematically. Specifically, for the first statement, I can explain it’s true because no matter how many times you multiply odd numbers, you will continue getting odd numbers, but this doesn’t seem very rigorous.

For the second statement, my proof is more rigorous as I say that

(2m)$^n$ = 2$^n$m$^n$, which must be even as 2$^n$ is always even.

Any comments regarding whether these proofs are valid would be great.

  • $\begingroup$ Because an odd number is just $1\text{ mod }2$ so we can raise it to any power and we still have $1^n\equiv1\text{ mod }2$. Similarly with even numbers, $0^n\equiv0\text{ mod }2$. $\endgroup$ Aug 2, 2019 at 17:18
  • $\begingroup$ The second one is fine. The first one isn't really a proof, because you are just restating the contention in different terms, but you can change it to a proof by induction easily. $\endgroup$
    – saulspatz
    Aug 2, 2019 at 17:18
  • $\begingroup$ Are you familiar with mathematical induction? Please give some context - where did you encounter the problem? In a course on proofs, or number theory, or discrete math? High school or university? $\endgroup$ Aug 2, 2019 at 17:26

2 Answers 2


An alternative to induction for your interest (although the induction proof is better in many ways). Consider any odd number $n=2k+1$ where $k=0,1,2,...$. And let $m$ be any positive integer. Then you want to guarantee that $$ n^m = (2k + 1)^m \mbox{ is odd}$$ Well we can use a famous result known as Binomial Theorem, which states that $$ (x + y)^m = \sum_{j=0}^{m} {m \choose k} x^{j} y^{m-j}, \qquad {m \choose k}=\frac{m!}{k!(m-k)!}$$ To see that $$ (2k + 1)^m = \sum_{j=0}^{m} {m \choose k} (2k)^{j} 1^{m-j} = 1 + 2 \left( \sum_{j=1}^{m} 2^{j-1} k^j \right)$$ which the right hand side is clearly odd


For the first statement, you can start by showing that the product of two odd numbers is always odd: $$(2n+1)*(2m+1) = 4nm+2(n+m)+1 = 2(2nm+n+m)+1.$$ Then, you can use induction to show the desired result.

By the above, if $n$ is odd, $n^2$ is odd. Thus, $n*(n^2)$ is odd, ...etc.

To show this rigorously:

  1. Show a base case.
  2. Assume that for some $p > 1$, $n^p$ is odd.
  3. Note that $n^{p+1} = n^p*n$, which is the product of two odd numbers, and hence odd.

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