<s>[INST] How does expanding $(2s+1)^2$ help us prove "$n^2$ even $\implies$ $n$ even"? How can $n^2 = (2s + 1)^2= 4s^2 + 4s + 1$ help in proving "If $n^2$ is even then $n$ is even"?
 [/INST] 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.
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