Macedonian Mathematical Olympiad

The product of three consecutive positive integers is divisible by $3$, and $9^n-7\equiv 2(\mathrm{mod}3)$, so we can conclude that $9^n-7$ cannot be written as a product of three or more consecutive positive integers.

Let $9^n-7=m(m+1)$ for some positive integer $m$. The last equation is equivalent with the equation $4\cdot 9^n-27=(2m+1{)}^2$, i.e. with the equation $4\cdot 9^n-(2m+1{)}^2=27$. Since $\{1,3,9,27\}$ are all positive divisors of $27$ and $$2\cdot 3^n+2m+1>2\cdot 3^n-2m-1,$$ these two cases are possible: $$1)\{\begin{array}{l}2\cdot 3^n+2m+1=27\\ 2\cdot 3^n-2m-1=1\end{array}$$ $$2)\{\begin{array}{l}2\cdot 3^n+2m+1=9\\ 2\cdot 3^n-2m-1=3\end{array}.$$ In the case 1) if we sum the equations we get $3^n=7$, which is impossible.

In the case 2) if we sum the equations we obtain $3^n=3$, so $n=1$. If we replace in the first equation $n=1$ we obtain $2\cdot 3+2m+1=9$, so $m=1$. It is clear that, $9^1-7=1\cdot (1+1)$.