Induction Proof, $n^2$ is even $\implies$ $n$ is even

Question

If $n^2$ is even $\Rightarrow$ $n$ is even.

Test: $n = 1$. $2^2 = 4$ is even $\Rightarrow$ $2$ is even.

Assuming: $P(k)$ is true with $k$ even.

Test: $P(k+2)$: $k+2$ is even.

I can't finish it. I do not know what to do.

Answer 1

$$(k+2)^{2}=k^{2}+4k+4$$

Suppose $(k+2)^{2}$ is even. Then, as $4$ and $4k$ are even, $k^2$ is forced to be even (if it were odd, equality could not hold above as odd does not equal even). Invoking $P(k)$, it follows $k$ is even and therefore $k+2$ is too.

Answer 2

You start off by assuming the inductive hypothesis. That is, assume that $k^2$ being even implies that $k$ is even. Now, we further assume that $(k+2)^2 = k^2+4k+4$ is even and we set out to prove that $k+2$ is even. Thus, for some integer, say $l$, we have that:

$$k^2+4k+4 = 2l$$

Solving this for $k^2$ gives:

$$k^2 = 2l-4k-4$$

Factoring a two out from the right hand side gives:

$$k^2 = 2(l-2k-2)$$

Thus, by definition, $k^2$ is even. So we can invoke the inductive hypothesis to say that this implies that $k$ is also even. So one can quickly see that this implies $k+2$ is even and the proof is complete.

Answer 3

HINT: Is $(n+2)^2$ even? Is the addition of even numbers even?

Answer 4

First, note that $1^2=1$ is odd. Now assume that we know for some odd $k$ that $k^2$ is odd. Then $(k+2)^2=k^2+4k+4$ is the sum of an odd term ($k^2$) and an even term ($4k+4$), so it too must be odd. You have now proved by induction that $k$ odd $\Rightarrow k^2$ is odd, which means that if $k^2$ is even, $k$ must also be even.

You can use a similar argument to prove that in fact any even number does have an even square (in other words, $k$ even $\iff k^2$ is even), but that's not necessary to prove what you've been asked to prove.

Answer 5

$2$ is prime, so one can apply Euclid's lemma, to get: $2|n^2\implies (2|n\lor2|n)\implies2|n$.

I don't see any need for induction.