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

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

Uniteasy Admin · Accepted Answer

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, $$n = 1963, 1993, 2004, 2011$$

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