A natural No 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.

• Let Squares are p2 and (p + 1)2
  given p^2 = n - k
  (p + 1)2 = n + l
  Now n- kl
  = n - (n - p2) ((p + 1)^2 - n)
  = n - n(p + 1)^2 + n^2 + p^2(p + 1)^2 - np^2
  = n - np^2 -2np - n + n^2 - np^2 + p^2(p + 1)^2
  = n2 - 2np^2 - 2np + p^2(p + 1)^2
  = n2 - 2np(p + 1) + p^2(p + 1)^2
  = [n - (p + 1)p]2
  so n-kL is a perfect square.
• 13 years agoHelpfull: Yes(7) No(2)