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 numberHence 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.