Saudi Arabian Mathematical Competitions, 2012

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

The triangles $BEC$ and $BE'A$ are congruent, so $BE' = BE$. Also, we have $\triangle BEF \cong \triangle BE'F$ by SAS congruency.

From $E'F = E'A + AF = EC + AF = (k - 12^{34}) + k - m$ and $E'F = EF = n$, it follows $n = (k - 12^{34}) + (k - m)$, and hence we get
$$k = \frac{1}{2}(m + n + 12^{34}). \quad (1)$$
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 $m \equiv n \pmod 2$, meaning $k$ is an integer. We can write
$$12^{68} = n^2 - m^2 = (n - m)(n + m) = 4 \cdot \frac{n - m}{2} \cdot \frac{n + m}{2}, \quad (2)$$
which is the same as
$$2^{134} \cdot 3^{68} = \frac{n-m}{2} \cdot \frac{n+m}{2}. \quad (3)$$
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.