Junior Balkan Mathematical Olympiad Shortlist

Solution:
Answer: $n=1$.

Among each four consecutive integers there is a multiple of $4$. As $36^{n}-6$ is not a multiple of $4$, it must be the product of two or three consecutive positive integers.

Case I. If $36^{n}-6 = x(x+1)$ (all letters here and below denote positive integers), then $4 \cdot 36^{n} - 23 = (2x+1)^2$, whence $(2 \cdot 6^{n} + 2x + 1)(2 \cdot 6^{n} - 2x - 1) = 23$. As $23$ is prime, this leads to $2 \cdot 6^{n} + 2x + 1 = 23$, $2 \cdot 6^{n} - 2x - 1 = 1$. Subtracting these yields $4x + 2 = 22$, $x = 5$, $n = 1$, which is a solution to the problem.

Case II. If $36^{n}-6 = (y-1)y(y+1)$, then
$$36^{n} = y^{3} - y + 6 = (y^{3} + 8) - (y + 2) = (y + 2)(y^{2} - 2y + 3)$$
Thus each of $y+2$ and $y^{2} - 2y + 3$ can have only $2$ and $3$ as prime factors, so the same is true for their GCD. This, combined with the identity $y^{2} - 2y + 3 = (y + 2)(y - 4) + 11$ yields $\operatorname{GCD}(y+2, y^{2} - 2y + 3) = 1$. Now $y+2 < y^{2} - 2y + 3$ and the latter number is odd, so $y+2 = 4^{n}$, $y^{2} - 2y + 3 = 9^{n}$. The former identity implies $y$ is even and now by the latter one $9^{n} \equiv 3 \pmod{4}$, while in fact $9^{n} \equiv 1 \pmod{4}$ — a contradiction. So, in this case there is no such $n$.