If n is positive integer greater than 1913, find the value of n such that \dfrac{n-1913}{2013-n} is a perfect square.

In order for \dfrac{n-1913}{2013-n} to be positive, n has to satisfy

1913 < n< 2013

Let m = n-1913

\dfrac{n-1913}{2013-n} = \dfrac{m}{100-m} =x^2

in which x is positive integers and m <100

Express m in terms of x,

m = \dfrac{100x^2}{1+x^2}

When x = 1,2,3,7, m = 50, 80, 90, 98, respectively

Accordingly, n = 1963, 1993, 2004, 2011