### 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.