ISI 2013 B.Math and B.Stat Subjective Solutions

1. For how many values of N (positive integer) N(N-101) is a square of a positive integer?

Solution:
(We will not consider the cases where N = 0 or N = 101)

$N(N-101) = m^2$

=> $N^2 - 101N - m^2 = 0$

Roots of this quadratic in N is
=> $\frac{101 \pm\ sqrt { 101^2 + 4m^2}}{2}$

The discriminant must be square of an odd number in order to have integer values for N.

Thus $101^2 + 4m^2 = (2k + 1)^2$
=> $101^2 = (2k +1)^2 - 4m^2$
=> $101^2 = (2k +2m + 1)(2k - 2m + 1)$

Note that 101 is a prime number

Hence we have two possibilities

Case 1:

$( 2k + 2m + 1 = 101^2; 2k - 2m + 1 = 1)$
Subtracting this pair of equations we get $(4m = 101^2 - 1)$ or $(4m = 100 \times 102$) or m = 50 × 51

This gives N = 2601 (ignoring extraneous solutions)

Case 2:

$(2k + 2m + 1 = 101 ; 2k - 2m + 1 = 101)$ which gives m = 0 or N = 101. This solution we ignore as it makes $N(N- 101) = 0$ (a non positive square).

Hence the only solution is $N = 2601$ and there are no other values of N which makes $N(N-101)$ a perfect square.