Question

the product of two consecutive natural numbers is 31 less than the sum of their squares. Find the number?​

Answer 1

Let N denote the smaller of the two natural numbers. Since the two numbers are consecutive, the larger natural number = N+1.

Given, Sum of the two numbers = 31

i.e., N+N+1 = 31

Or, 2N + 1 = 31

Subtracting 1 from both sides,

2N+1–1 = 31–1

Or, 2N = 30

Dividing both sides by 2,

2N/2 = 30/2

Or, N = 15 ………………………………………………………………..(1)

Method 1:

Difference of the squares of the two numbers

=(N+1)² - N² = N² + 2N + 1 - N² = 2N + 1

= 2 x 15 +1 (Substituting for N = 15 from (1))

=30+1 = 31 (Answer)

Method 2:

N = 15

∴ N+1 = 16

and N² = 15² = 225, (N+1)² = 16² = 256

∴Difference of their squares

= (N+1)² - N² = 256 - 225 = 31 (Answer)

Theorem:

It is interesting to observe that the sum of the two consecutive natural numbers is equal to difference of their squares. This is true for every such pair of natural numbers if:

(I) They are consecutive

(ii) If the sum is an odd number?

Therefore, number of such numbers is infinite. As example:

29, 30

Sum= 30+29=59

Difference of squares =

30^2 - 29^2 = 900–841 = 59