Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.
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 $4 \cdot 36^{n}-23=(2 x+1)^{2}$, whence $\left(2 \cdot 6^{n}+2 x+1\right)\left(2 \cdot 6^{n}-2 x-1\right)=23$. As 23 is prime, this leads to $2 \cdot 6^{n}+2 x+1=23,2 \cdot 6^{n}-2 x-1=1$. Subtracting these yields $4 x+2=22, x=5, n=1$, which is a solution to the problem.

Case II. If $36^{n}-6=(y-1) y(y+1)$, then

$$
36^{n}=y^{3}-y+6=\left(y^{3}+8\right)-(y+2)=(y+2)\left(y^{2}-2 y+3\right)
$$

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