A letter about odd and even

Dear friend, I found a trick today. Take any two odd numbers and add them: the answer is always even. Three plus five is eight, seven plus nine is sixteen.

Dear friend, I found a trick. For odd $a$ and $b$, $a+b$ is even: $3+5=8$ and $7+9=16$.

$a\equiv b\equiv 1(\mathrm{mod}2)\Rightarrow a+b\equiv 0(\mathrm{mod}2)$ and $3+5=8,7+9=16$.

Here is why. An odd number is an even number plus one. So two odd numbers are two even numbers plus one and one, and one and one make a pair. But odd times odd is different: it stays odd. Three times five is fifteen.

Why: odd means $a=2m+1$ and $b=2n+1$, so $a+b=2(m+n+1)$ is even. But $ab=2(2mn+m+n)+1$ is odd: $3\times 5=15$.

$(2m+1)+(2n+1)=2(m+n+1)$ and $(2m+1)(2n+1)=2(2mn+m+n)+1$.

Now the surprise. Add the odd numbers from the start, one after another. One is one. One plus three is four. Plus five makes nine, plus seven makes sixteen. One, four, nine, sixteen: these are the squares. The sum of the first n odd numbers is n times n.

The surprise: $1=1,1+3=4,1+3+5=9,1+3+5+7=16$. These are the squares, and ${\sum }_{k=1}^n(2k-1)=n^2$.

$1=1^2,1+3=2^2,1+3+5=3^2,1+3+5+7=4^2$ and ${\sum }_{k=1}^n(2k-1)=n^2$.

Before you write back, try the first ten odd numbers. You should get one hundred. Yours, with a pencil in my hand.

Try the first $10$ odd numbers: you should get $10^2=100$. Yours, with a pencil.

${\sum }_{k=1}^{10}(2k-1)=10^2=100$.