Answer to Question #13400 in Algebra for pankaj bhist

Question

natural number n is chosen strictly between two consecutive perfect squares.
The smaller of these two squares is obtained by subtracting k from n and the
larger one is obtained by adding l to n. Prove that n − kl is a perfect square.

Answer 1

Let the number n be between a² and (a+1)². As per given condition

n-a² = k,& (1)
(a+1)² –n = l.& (2)

Adding (1) and (2):

k + l = 2a + 1,

or

l = 2a + 1 - k.

Now

n-kl = (a²+k) – k(2a+1-k)
= a² + k – 2ak –k + k²
= a² – 2ak + k² = (a-k)².

Hence proved that n-kl is a perfect square.