Odd Squares

By Sebastiaan

No, this is not a post about strange squares. It's all about odd squares though.

Warning! Non-mathematicians may find this post unreadable.

Let me list the first few:

1² = 1
3² = 9
5² = 25
7² = 49
9² = 81

At first glance, there does not seem to be a pattern. I got this puzzle from my school, where I study to be a mathematics teacher. It stated:

"The squares of all positive odd integers are eightfolds plus one."

First thing to do was translate this into proper mathematics language.

So, odd, would mean the integer can be written as "2k+1" with k any positive integer. Thus we get:

(2k+1)²

Also, an eightfold, is "8n" with n any positive integer. Thus we get:

(2k+1)² = 8n + 1

Now let's work out the brackets...

4k² + 4k + 1 = 8n + 1
4k² + 4k = 8n
for all positive integers k and n.

So now we have a relatively simple statement to proof. First thing that springs to mind is induction, but if you look just a little bit further... you'll find a much simpler and more elegant proof.

So, let's first start by dividing by four.

k² + k= 2n

What's left to proof is that the sum of an integer and its square is even. There are multiple approaches to solving this, I'll give you two.

Say, k is even. Then k² is most certainly even, for (2p)² is 4p² which is a multiple of four and thus even. When you add another even integer, it will always remain even.
Say, k is odd. Then k² is odd, for (2p+1)² is 4p² + 4p + 1, which are two even integers plus one, making it odd. Now add the same odd integer k, you will get an even integer, for two odds make an even.

So, for any positive integer k, we get an even integer. Thus every odd square is an eightfold plus one.
Q.E.D.

Another approach could be rewriting the left-hand of the equation, that would make k(k+1) = 2n. Now, an integer multiplied by the next integer, would always be 'odd' times 'even' or 'even' times 'odd'. This always makes even.
Q.E.D.