number theory 2013

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

that is $( N^2 – 101N – m^2 = 0 $)

Roots of this quadratic in N is
that 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 $)
that is $( 101^2 = (2k +1)^2 – 4m^2 $)
that is $( 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 $\cdot 51$

This gives N = 2601

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

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