Saturday, 29 December 2012

Calculations made easier...

If calculating squares was a tedious job, calculating the difference of squares can be an even tougher task especially if there are bigger numbers involved. 
However, there is a neat trick that we have found which can produce the desired answer without actually calculating the squares.

There are two approaches, one is the one developed by us and the other one is more conventional, but equally effective.

  • ·       The 1st method involves generalizing the equation of differences of squares:  Let ‘n’ be the greater number and ‘n-d’ be the smaller number where‘d’ is the difference between these two numbers. Then, (n) 2-(n-d) 2 = n2-(n2+d2-2dn) = 2dn-d2. Hence we can express the difference between two squares via this simple expression (2dn-d2). The best thing is that now we need not calculate the squares of the bigger numbers involved, we just have to calculate the square of the difference which would be much smaller. 

Eg. 212-122 (n=21 d=9) Using 2dn-d2
2(9) (21)-(9)2=378-81=297

In case there is a negative number involved, one need not consider the negative sign for the difference, because square of a negative number is a positive number.
(15)2-(7)2 = (15)2-(-7)2

  • ·       The 2nd method makes use of the identity [a2-b2= (a+b) (a-b)]. This method too saves the hassle of calculating the squares individually and is especially useful if there are small differences involved.

Eg. 802-782= (80+78) (80-78) =158x2=316

Squaring numbers-The better way!! contd..

This article is an extension to our fist article: ‘Squaring numbers-The better way!!’ We will demonstrate how to calculate squares using the simple identity-


This method is useful as it helps to break up the number into a multiple of 10 whose square is pretty use to find.
Eg. (76)2= (70+6)2=702+2(70)(6)+62=4900+840+36=5776

By splitting 76 into 70 and 6, we were able to calculate its square quickly, by simply using the identity.

Friday, 21 December 2012


Even though there exist infinitely many Pythagorean triplets, the most unique ones are those originating from 3, 4 and 5. 

It is known that:
32 +42 =52

We can generalize these triplets in the form:
(3n) 2+ (4n) 2= (5n) 2 

Where ‘n’ is any whole number >0.

Let 3n=a  4n=b  5n=c 

It is observed that (c/a) = (5/3) = φ ≈ 1.66= Golden Ratio

Using the above formula, we can use various values of ‘n’ to find infinitely many such ‘golden triplets’

We have named them golden triplets because they follow the golden ratio.

For information about the golden ratio, please visit

Wednesday, 19 December 2012

Squaring numbers-The better way!!

Calculating the squares of large numbers can be a difficult task but we have found some relations between the numbers that can easily reduce the calculation time and help evaluating the squares

n2 = (n+1) (n-1) +1

i.e  Square of a number= (Preceding number)(Succeeding Number)+1

For e.g.:- n=99
992= (99-1) (99+1) + 1
      = 98*100 + 1
      = 9801

Hence, now we know that square of 99 = 9801

The other relation is:

n2 = (n-1)2 + (n-1) + n

i.e square of a number= Square of preceding number+Preceding Number+The number itself

For e.g.:- n=13
132 = (13-1)2 + (13-1) +13
       = 144 +25
       = 169

We can combine the above two identities to form another useful relation

Suppose, we have n=98
We know that 992 = 9801 [i.e. (n+1)2 =9801] from the identity no.1. So, by using the identity no.1 and no.2 we form a relating:-

(n+1)= n2 +n + (n+1)

Hence, we can conclude that,
      n2 = (n+1)2 - n – (n+1)

By putting n=98
             = (99)2 - 98 - 99
            = 9801 - 98 - 99
            = 9604