Square of Odd Numbers Minus 1

Many numbers have interesting properties worth investigating. These properties can be used to introduce the notion of proofs. Teachers may try this out in middle school or high school students. Although the proofs might not be as elegant or as formal such as written in books, its notion and the reasoning behind it is more important. Let us try to answer the question below.

Investigate the numbers which are one less than the square of odd numbers.

We can try some examples and see their properties.

3^2 - 1 = 8

5^2 - 1 = 24

7^2 - 1 = 48

9^2 - 1 = 80

Looking at the pattern, it is quite obvious that all of them are divisible by 8. Maybe, you might want to have more examples. To lessen the burden the calculation, you might want to use a spreadsheet to do this. 


We have already learned how to represent even numbers and odd numbers. Even numbers can be represented by 2k, while odd numbers can be represented by 2k + 1. Thus, we can generalize the conjecture above.


The square of odd numbers minus 1 is divisible by 8.


Odd numbers can be represented as 2k + 1 where k is an integer. So, the square of an odd number minus 1 can be represented by

(2k + 1)^2 - 1

Notice that the expression above is a difference of two squares. We can see this easily if we let 2k + 1 be equal to n. Substituting n to the equation above, we have

(2k + 1)^2 - 1 = n^2 - 1.

Factoring the difference of two squares, we have $latex n^2 – 1 = (n + 1)(n-1).

Substituting back the value of 2k + 1 to n, we have

(n + 1)(n – 1) = (2k + 1 + 1)(2k + 1 – 1)

The right hand side of the equation and factoring out 2 results to

(2k + 2)(2k) = 2(k + 1)(2k) = 4k(k+1).

Now, since 4k(k+1) has 4 as factor, it is divisible by 4. Now, k and k + 1 are consecutive integers which means that one of them is even (why?). If one of them is even, then 2 is one of its factor.  This means that 4k(k+1) has factors 4(2) = 8. If it has 8 as a factor, then it is divisible by 8. This is what we want to prove.

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