Is this the junior Olympiad?

Let $9^n-7 = a(a+1)...(a+k)$. Note that the RHS is divisible by $(k+1)!$ but $9^n-7$ is never divisible by $3$ (unless $n=0$ but it will be negative) so we get that $k=1$. Now \begin{align*} &4\cdot 9^n - 28 = 4a^2 + 4a \\ &\implies (2\cdot 3^n)^2 - (2a+1)^2 = 27 \\ &\implies (2\cdot 3^n + (2a+1))(2\cdot 3^n - (2a+1)) = 27 \end{align*}so the two factors are $\{1,27\}$ or $\{3,9\}$ which must sum to $4\cdot 3^n$. So the only possible way is $n=1$ which we conclude that $n=1$ are all possible solutions.

MarkBcc168 wrote: Let $9^n-7 = a(a+1)...(a+k)$. Note that the RHS is divisible by $(k-1)!$ but $9^n-7$ is never divisible by $3$ (unless $n=0$ but it will be negative) so we get that $k=1$. Now \begin{align*} &4\cdot 9^n - 28 = 4a^2 + 4a \\ &\implies (2\cdot 3^n)^2 - (2a+1)^2 = 27 \\ &\implies (2\cdot 3^n + (2a+1))(2\cdot 3^n - (2a+1)) = 27 \end{align*}so the two factors are $\{1,27\}$ or $\{3,9\}$ which must sum to $4\cdot 3^n$. So the only possible way is $n=1$ which we conclude that $n=1$ are all possible solutions. Nice MarkBcc168!

Hamel wrote: Is this the junior Olympiad? No, senior olympiad.

MarkBcc168 wrote: so we get that $k=1$ Why can $k$ not be $2$ or $3$? Sorry if the question is dumb.

Vrangr wrote: MarkBcc168 wrote: so we get that $k=1$ Why can $k$ not be $2$ or $3$? Sorry if the question is dumb. I think they meant $\left(k+1\right)!$

Vrangr wrote: MarkBcc168 wrote: so we get that $k=1$ Why can $k$ not be $2$ or $3$? Sorry if the question is dumb. When $k \geq 2$ $a(a+1)(a+2)...(a+k)$ divisible by 3 but $9^n-7$ is never divisible by 3

steppewolf wrote: Determine all natural numbers $n$ such that $9^n - 7$ can be represented as a product of at least two consecutive natural numbers. I am going to solve the folowing :Determine all natural numbers $n$ such that $3^n - 7$ can be represented as a product of at least two consecutive natural numbers. If it is a pruduvt of at leat tree then $3|7$ contradiction. so $3^n-7=k(k+1)$ $4*3^n-27=(2k+1)^2$ for $n<4$ we have solution only for $n=2,3$ if $n>3$ then $27|LHS$ which means that $81|RHS$ which leads to contradicton as $81$ does not devise $27$.

First of all if it could be represented as a product of three natural numbers, then it would be divisible by $3$ which is untrue. Next putting $9^n -7=m^2+m$ and checking discriminant to be a perfect square, we have that $n<4$ so now just case checking finishes.