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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.

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