How does expanding $(2s+1)^2$ help us prove “$n^2$ even $\implies$ $n$ even”?

Question

How can $n^2 = (2s + 1)^2= 4s^2 + 4s + 1$ help in proving "If $n^2$ is even then $n$ is even"?

Answer 1

Based on your comment on Lazar Ljubenović answer, here is an attempt to explain how to expand $(2s+1)^2$.

First let us rewrite $(2s+1)^2$ as $(2s+1)(2s+1)$. For the sake of this explanation let us rewrite $(2s+1)(2s+1)$ as $c(2s+1)$ where $c=2s+1$.

Using the distributive law $a(b+c)=ab+ac$ we know $$c(2s+1)=c\cdot 2s+c\cdot 1=c\cdot 2s+c$$

Now if we substitute back $c$ we get

$$2s\cdot (2s+1)+(2s+1)$$

If we use the distributive law once again we get:

$$(2s)\cdot (2s)+2s\cdot 1+2s+1=2\cdot2\cdot s\cdot s+2s+2s+1$$ Which is

$$4s^2+4s+1$$

As for the question: how can this help prove that if $n$ is odd, $n^2$ is also odd?

Since any even number plus 1 is odd, any odd number can be written in the form $2n+1$ where $n$ is an integer. If we square $2n+1$ we get $(2n+1)^2=4n^2+4n+1$. Note that this can be simplified to $4(n^2+n)+1$. Since $4(n^2+n)$ is always even, $4(n^2+n)+1$ must be odd.

Answer 2

Hint $\ $ If $\rm\:n\:$ is odd then $\rm\:n = 2s+1,\:$ so expanding $\rm\:n^2 = (2s+1)^2\:$ shows $\rm\: n\ odd\:\Rightarrow\:n^2\ odd,\:$ or, contrapositively, $\rm\:n^2\ not\ odd\:\Rightarrow\: n\ not\ odd,\:$ i.e. $\rm\:n^2\ even\:\Rightarrow\:n\ even.$

Answer 3

If I understand your question correctly:

Let $n=2s+1$. That's and odd number because it's greater than even number ($2s$) by one. So, when you square it, you get: $(2s+1)^2=4s^2+4s+1=4(s^2+s)+1$. This is also an odd number, because it's a number you get when you add $1$ to even number $4(s^2+s)$.

With this, your proof that if $n$ is odd, then $n^2$ is odd as well is done.

Answer 4

An odd number is a number of the form $(2\cdot\text{something})+1$.

So the square of an odd number is $$(2s+1)^2 = 4s^2+4s+1 = 2\Big(2s^2+2s\Big) + 1 = (2\cdot\text{something})+1.$$

Therefore the square of an odd number is odd.

Beyond that, you need to know that every integer is either odd or even.

Answer 5

Michael Hardy has the key observation: "you need to know that every integer is either odd or even."

On the Peano level, this seems non-trivial. My attempt is this (where $s$ is the successor function):

Base case: 1 is odd; s(1) is even.

Induction: If n is even, s(s(n)) is even; if n is odd, s(s(n)) is odd.

It seems to me we have to show that (1) every integer is even or odd; (2) no integer is both even or odd. To do this, we have to show that (1) if n is even, s(n) is odd; if n is odd, s(n) is even.

Then, I think, we can prove these by looking at the smallest integer that is neither even or odd and getting a contradiction; the same for the smallest integer that is both even and odd.

Am I making too much of this, or is there a simple way to do this?