Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even
the proof at hand is:
Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even.
Hint: Contradiction
So I know to start the problem in a contradiction format would be:$a^n$ is even, then $a$ is odd so that $a = 2k+1$. Then plug in that into $a^n$, we get $(2k+1)^n$. This is where I become stuck.
Thanks in advance!
$\endgroup$ 36 Answers
$\begingroup$Hint: expand using the binomial theorem.
$\endgroup$ 2 $\begingroup$$$(2k+1)^n=2q+1\quad,\quad \text{for some }q\in\Bbb Z$$since$$(2k+1)^n{=\sum_{i=0}^n(2k)^i\\=1+\sum_{i=1}^n(2k)^i\\=1+2\cdot\sum_{i=1}^n2^{i-1}\cdot k^i\\=1+2q}$$
$\endgroup$ 5 $\begingroup$By Euclid's lemma, if $2|a^n$ then $2|a$ or $2|a$ or $2|a$ or ... or $2|a$,
so we conclude that $2|a$.
$\endgroup$ $\begingroup$The contrapositive is that $$a \text{ odd }\implies a^n \text { odd}$$
Use induction here: if $a^1=2k_1+1$, then $a^2=(2k_1+1)^2=2(2k_1^2+2k_1)+1=2k_2+1$hence true for $n=2$
Then assume true for $n=m$, $(2k_1+1)^m=2k_m+1$
Then $$(2k_1+1)^{m+1}=(2k_m+1)(2k_1+1)=4k_mk_1+2k_m+2k_1+1=2(2k_mk_1+k_m+k_1)+1$$
Hence true with $k_{m+1}=2k_mk_1+k_m+k_1$
$\endgroup$ $\begingroup$You can prove this by induction:
For $n=1$ the claim is clear.
Assume that if $a^n $ is even then $a $ is even.
Let $a^{n+1}$ be even. Then:
$a^{n+1} = a a^n$ .
If a is not even, then $a^n$ must be even because otherwise their product won't be even.
But then , by the induction hypothesis , $a$ is even, contradiction.
So $a$ must be even, as wanted.
$\endgroup$ $\begingroup$Prove by contradiction : If $a^n$ is even then a is an odd number.
We have : a = 2k + 1 By assumption above : $(2k+1)^n$ = 2q for all q $\in$ Z
Using binomial theorem we have : $(2k+1)^n$ = $\sum_{i = 0}^n C_i^n(2k)^i$ = 1 + 2$\sum_{i = 1}^n C_i^n(2k)^{i - 1}$.
We can derive that q is a rational number contradicting the fact that q is an integer.
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