Saudi Arabia Mathematical Competitions 2012

Let $\widehat{DBE}=x$, $\widehat{DBF}=y$, and construct the point $E^{\prime }$ on the ray $DA$ such that $\widehat{E^{\prime }BA}=y$.



The triangles $BEC$ and $B{E}^{\prime }A$ are congruent, so $B{E}^{\prime }=BE$. Also, we have $△BEF\cong △B{E}^{\prime }F$ by SAS congruency.

From $E^{\prime }F=E^{\prime }A+AF=EC+AF=(k-12^{34})+k-m$ and $E^{\prime }F=EF=n$, it follows $n=(k-12^{34})+(k-m)$, and hence we get $$k=\frac{1}{2}(m+n+12^{34}).(1)$$ The triples $(k,m,n)$ are in bijection with the pairs $(m,n)$ satisfying $m^2+12^{68}=n^2$. Indeed, any integral solution $(m,n)$ to $m^2+12^{68}=n^2$ satisfies $m\equiv n(\mathrm{mod}2)$, meaning $k$ is an integer. We can write $$12^{68}=n^2-m^2=(n-m)(n+m)=4\cdot \frac{n-m}{2}\cdot \frac{n+m}{2},(2)$$ which is the same as $$2^{134}\cdot 3^{68}=\frac{n-m}{2}\cdot \frac{n+m}{2}.(3)$$ For every divisor $d$ of $2^{134}\cdot 3^{68}$, with $d^2<2^{134}\cdot 3^{68}$, we obtain a solution $\frac{n-m}{2}=d$, $\frac{n+m}{2}=\frac{2^{134}\cdot 3^{68}}{d}>d$ and conversely. We get $\frac{135\cdot 69-1}{2}$ solutions since $2^{134}\cdot 3^{68}$ has $135\cdot 69$ divisors.