Let N be an integer such that N(N – 101) is a perfect square of a positive integer .

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

that is )

Roots of this quadratic in N is

that is )

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

Thus )

that is )

that is )

Note that 101 is a prime number

Hence we have two possibilities

Case 1:

)

Subtracting this pair of equations we get or or m = 50

This gives N = 2601

Case 2:

) 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 for which N(N-101) is not a perfect square .

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