# Answer to Question #13400 in Algebra for pankaj bhist

Question #13400

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.

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.

Expert's answer

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.

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.

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