### The Problem

### Choose four consecutive odd counting numbers. Take the product
of the middle two numbers and subtract the product of the first
number and the last number. Try a few samples and formulate a
rule. Explain why the rule works.

#### The Solution

First, let's look at a few examples to see if we can notice
any sort of pattern.

Ex. 1) 31, 33, 35, 37

(33 x 35) - (31 x 37)

1155 - 1147 = 8

Ex. 2) 9, 11, 13, 15

(11 x 13) - (9 x 15)

143 - 135 = 8

Ex. 3) 1233, 1235, 1237, 1239

(1235 x 1237) - (1233 x 1239)

1,527,695 - 1,527,687 = 8

This is interesting, we seem to be getting a consistent answer
of 8. Now we can try to create a generalization of the pattern.
Lets call the odd number we begin with n. Then the next odd number
would be n+2, followed by n+4, and finally n+6. So we can write
the following generalization for any odd number n.

Ex. 4) n, n+2, n+4, n+6

(n + 2) (n + 4) - (n) (n + 6)

(n^2 +6n +8) - (n^2 +6n) = 8

We can see that for any four consecutive odd integers, subtracting
the product of the first and last numbers from the product of
the middle two numbers will always give a difference of 8.

###### An Extension to Even Integers

The above rule will hold true for consecutive even integers
as also; if we pick any even integer n, the next even integer
would be n+2, allowing consecutive even integers to follow the
above rule.

The rule will change slightly for consecutive integers. Again
we can pick any integer n, but the next three integers will be
n+1, n+2, and n+3. Thus the rule changes in the following manner:

Ex. 5) n, n+1, n+2, n+3

(n +1) (n +2) - (n)(n + 3)

(n^2 + 3n +2) - (n^2 + 3n) = 2

So, taking four consecutive integers and applying the perimeters
of the above problem will result in a difference of 2.