Prove that if $n^k$ is even, then $n$ is an even integer.

Question

I proved the converse (if $n$ is an even integer, then $n^k$ is even) using induction.

Inductive step: $n^{k+1}=n\cdot n^k$. Since it is assumed $n^k$ is even, an even times an even gives an even number.

But I am not sure how to go about proving if $n^k$ is even, then $n$ is an even integer. Any help is greatly appreciated.

Thanks!

Answer 1

If $n^k$ is even it means $2|n^k$. This means that $2$ divides one of the factors of $n^k$ since all the factors are equal it means that $2|n$. In other words if $2$ divides $n^k$ then one of the factors must be even, therefore $n$ must have the form $2l$.

Answer 2

If $n$ is not even, then it must be odd. However, any power of an odd number is again odd. [Hint: you may want to try the binomial expansion of $2k+1$]

Contradiction.

Hence $n$ is even.

Answer 3

If $n \equiv 1 \pmod{2}$, then $n^k \equiv 1 \pmod{2}$. That is if $n$ is odd, then $n^k$ must be odd.

Hence, if $n^k$ is even, $n$ must be even.

Answer 4

By contrapositive:

Suppose $n$ is not even, i.e. $n$ is of the form $n=2m+1$ for some integer $m\geq 0$.

Then, by the binomial theorem, $$ n^k = (2m+1)^k = \sum_{\ell=0}^k \binom{k}{\ell} (2m)^\ell = 1+\sum_{\ell=1}^k \binom{k}{\ell} 2^\ell m^\ell\,. $$ Now, every $\binom{k}{\ell}$ is an integer, and so is $m^\ell$: meaning that for $\ell\geq 1$ we have that $2^\ell\cdot \binom{k}{\ell} m^\ell$ is even. This implies that $\sum_{\ell=1}^k \binom{k}{\ell} 2^\ell m^\ell$ is even as sum of even integers, and therefore that $1+ \sum_{\ell=1}^k \binom{k}{\ell} 2^\ell m^\ell$ is odd.

So if $n$ is not even, then $n^k$ is not even.

Answer 5

Use the same argument to prove that if $n$ is odd, then $n^k$ is odd.

Its contrapositive is: if $n^k$ is even, then $n$ is even

Answer 6

For fun:

Let $k \in \mathbb{Z^+}.$

Prove by induction for $k$:

If $n^k$ is even then $n$ is even.

1) $k=1$, is ok.

2) Hypothesis:

The statement is true for $k$.

3) Step :

Consider $n^{k+1}= n \cdot n^k$, where

$n^{k+1}$ is even:

Then: $2| n^{k+1}$ .

Euclid's lemma:

If $p$, prime, divides $ab$, then $p|a$ or $p|b$.

Hence : $2| n$ or $2|n^k.$

1) If $p|n$ we are done .

2) If $p|n^k$ we are done by hypothesis .