2011/106) A natural number n is chosen strictly between two consecutive perfect squares.

The smaller of these two squares is obtained by subtracting k from n an the larger is obtained by adding l to n. Prove that n-kl is a perfect square.

Proof:

Let the number n be between a^2 and (a+1)^2

As per given condition

n- a^2 = k …1

(a+1)^2 –n = l ….2

Adding (1) and (2)

k + l = 2a + 1

or l = 2a + 1 - k

now

n- kl = (a^2+k) – k(2a+1-k)

= a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2

Hence proved

that n-kl is a perfect square.