26th Hellenic Mathematical Olympiad

It is enough to prove that there exist $a,b\in {\mathbb{N}}^{\ast }$ with $(a,b)=1$ such that: $$\frac{9n-1}{n+7}=\frac{a^2}{b^2}(1)$$ From this relation we get: $$n=\frac{7a^2+b^2}{9b^2-a^2}=\frac{7(a^2-9b^2)+64b^2}{9b^2-a^2}=-7+\frac{64b^2}{9b^2-a^2}(2)$$ Since $(a,b)=1$, it follows that $(a^2,b^2)=1$ and $(9b^2-a^2,b^2)=1$, and hence from (2) we get that $n$ is integer, if and only if $9b^2-a^2$ is a divisor of 64. Since $a,b$ and $n$ are positive integers, it follows that $9b^2-a^2\ge 8$, and hence: $$9b^2-a^2=(3b+a)(3b-a)\in \{8,16,32,64\}.(3)$$ Moreover, the factors $3b+a$, $3b-a$ have sum a multiple of 6 and difference a multiple of 2 and $3b+a>3b-a$. Therefore from relation (3) the possible cases are the following: $$\begin{array}{rl}(3b+a,3b-a)& =(4,2)\eta (3b+a,3b-a)=(8,4)\eta (3b+a,3b-a)=(16,2)\\ \iff (a,b)& =(1,1)\eta (a,b)=(2,2)\eta (a,b)=(7,3).\end{array}$$ The pair $(a,b)=(2,2)$ is rejected, because $\text{gcd}(2,2)=2\ne 1$, and therefore we have the values $n=1$ or $n=11$.