I am a frequent victim of insomnia. Instead of counting sheep, I often entertain myself with little mathematical exercises until I finally drift off. Here is one from a recent sleepless night. I had fun with it and thought you might find it interesting…and an opportunity to brush-up on your high school algebra.

2 x 4 = 8

3 x 3 = 9

3 x 5 = 15

4 x 4 = 16

4 x 6 = 24

5 x 5 = 25

Notice that the product of two numbers differing by two is one LESS than the square of the intervening number. Did you know that? I didn’t. Is it always true?

A little algebra gives the answer:

let x = 2

then 2 x 4 can be written as x(x+2) = x^{2} + 2x

and 3 x 3 can be written as (x+1)^{ 2} = x^{2} + 2x + 1

So, the product of any two numbers differing by two will always be one less than the square of the intervening number.

You can use this to wow your friends. Ask them what 21 x 19 equals. When they look blank, just say that it’s 399 and walk away. Or 99 x 101. That’s so easy…9999. How about 24 x 26? 624.

This can be extended to products of two numbers differing by 4 instead of two. In this case:

x(x+ 4) = x^{2} + 4x

The square of the midpoint value (x + 2) is:

x^{2} + 4x + 4

So, the midpoint value squared is 4 greater than the product of two numbers differing by 4.

Example:

2 x 6 = 12

4 x 4 = 16

Products of number pairs that differ by an odd number are also interesting:

2 x 5 = 10

What is the product of the intervening two numbers?

3 x 4 = 12

algebraically:

x(x + 3) = x^{2} + 3x

(x + 1)(x + 2) = x^{2} + 3x + 2

For two numbers differing by five, the product of the __middle__ intervening pair will always be greater by 6.

There are two general rules that can be derived from this.

The product of any two integers differing by an integer n is:

x(x+n) = x^{2} + nx

When n is an even integer, the square of the midpoint value is:

(x+n/2)^{ 2} = x^{2} + nx + (n/2)^{ 2}

__RULE 1__: The product of any pair of numbers differing by an __even__ integer n will be less than the square of the midpoint value between them by n^{ 2}/4.

When n is an odd integer, the product of the midpoint pair of numbers is:

(x + (n – 1)/2)(x + (n + 1)/2) = x^{2} + x(n + 1)/2 + x(n – 1)/2 + ((n-1)(n+1))/4

This reduces to x^{2 }+ nx + (n^{2 }– 1)/4

__RULE 2__: The product of any pair of numbers differing by an __odd__ integer n will be less than the product of the midpoint pair between them by (n^{2} – 1)/4.

The most surprising thing to me is that the difference in the products is independent of the value of x. That is not what my intuition would have told me.

Thinking about this did help me get to sleep. I hope that reading about it didn’t put you to sleep.

*Bert Bigelow graduated from the University of Michigan engineering school, and then pursued a career in software design. He has always enjoyed writing, and since retirement, has produced short essays on many subjects. His main interests are in the areas of politics and religion, and the intersection of the two. Many of his writings are posted on his web site, bigelowbert.com. You can contact him at bigelowbert@aol.com.*