Greece National Olympiad Problem 2

$A = \sqrt{\frac{9n-1}{n+7}}$

Now we know that $A = \frac{b}{c}$

Then $\frac{9n-1}{n+7} = \frac{{b}^{2}}{{c}^{2}}$

And so $9n - 1 = a{b}^{2}$

$n + 7 = a{c}^{2}$

or $9n+63 = a{(3c)}^{2}$ and $9n - 1 = a{b}^{2}$

And so -

$64 = a(3c - b)(3c + b)$

Now we can easily find solutions as we have to find solutions such that $3c - b = {2}^{x}$ and $3c + b = {2}^{y}$ and so $6|({2}^{x}+{2}^{y})$ , which are easy to find(by simple arithmetical calculations)

Solutions $(a,b,c)$ as $(8,1,1), (2,7,3), (2,2,2)$ [taking out negative case]

We want to find $n$ and $n = a{c}^{2} - 7$ .

$n = 8*({1}^{2}) - 7 = 1$

$n = 2*({3}^{2}) - 7 = 11$

$n = 2*({2}^{2}) - 7 = 1$

And so $n = 1, 11$ . Answer is $1 + 11 = 12$