# Expressing a number as a difference of two squares

Any whole number can be expressed as a product of two whole numbers. If the number is prime, it can be expressed as a product of itself and 1. The algebraic formula a^{2} – b^{2} = (a + b)(a – b) shows that the product of the sum (a + b) and their difference (a – b) **is** a difference of two squares.

Case **A **is easy to illustrate. 17, a prime number can be expressed only as a product of 17 and 1. So let (a – b) = 1; that is, let **a **and** b **be consecutive numbers adding up to 17. **a** and **b **are thus** 9 **and** 8 **respectively**.**

Therefore, ( 9 + 8) ( 9 – 1) = (9^{2} – 8^{2}) = 81 – 64 = 17

In case **B, **24 can be expressed as 24 x 1, 12 x 2 , 8 x 3 and 6 x 4. We can immediately see that we can not apply the method we used in case **B**. We need a more general solution. In his book *Vedic Mathematics*, Sri Bharati Tithaji showed how we can do that.

If we expand (a + b)^{2} and (a – b)^{2} and take their difference, we would get

(a^{2} + 2ab + b^{2}) – (a^{2} – 2ab + b^{2}) = 4ab or

(a + b)^{2} – (a – b)^{2} = 4 ab and

^{2}– [(a – b)/2]

^{2}= ab

Now if we take 24 as 12 x 2, we would have (12 + 2)/2 = 7 and (12 – 2)/2 = 5 and

7^{2} – 5^{2 }= 49 – 25 = 24.

For 24 = 6 x 4, we have (6 + 4)/2 = 5 and (6 – 4)/2 = 1 ; 5^{2} – 1^{2} = 25 – 1 = 24.

We can also apply it for 3 x 8. (8 + 3)/2 = 11/2 and (8 – 3)/2 = 5/2;

(11/2)^{2} – (5/2)^{2} = 121/4 – 25/4 = 96/4 = 24

For 1 x 24, we have (24 + 1)/2 = 25/2 and (24 – 1)/2 = 23/2;

(25/2)^{2} – (23/2)^{2} = 625/4 – 529/4 = 96/4 = 24

Let us now apply this formula in case **B**. (17 + 1)/2 = 9 and (17 – 1)/2 = 8. We will again have 9^{2} – 8^{2} = 81 – 64 = 17.

Later we will see how this formula can be used in determining the sides of a right triangle.

Suggested readings: Sri Bharati Tirthaji, Vedic Mathematics, pp 281-284.