Why is powers of 6 scared of powers of 7?

Note that the left-hand side is obviously divisible by $9$ .

What about the right-hand side? Use the formula for the sum of a geometric series.

$\begin{aligned} 6+6^2+\cdots+6^n&=\frac{6\left(6^{n+1}-1\right)}{6-1}\\ &=\frac{6\left(6^{n+1}-1\right)}{5} \end{aligned}$

Now, there's only a single factor of $3$ as $6^{n+1}-1$ is not divisible by $3$ . Thus, the right-hand side is never divisible by $9$ .

Note

As user X X pointed out in the comments, there is an easier way to show that the right-hand side is not divisible by $9$ :

$\begin{aligned} 6+6^2+\cdots+6^n&=6+36\left(1+6^2+\cdots+6^{n-2}\right)\\ &=6+9\cdot 4\left(1+6^2+\cdots+6^{n-2}\right) \end{aligned}$

We have that $\dfrac{ 9+9^2+\cdots +9^m}{6+6^2+\cdots +6^n} = 1 \\ \left(\dfrac{9\,(9^m-1)}{8}\right) \cdot \left(\dfrac{5}{6\,(6^n-1)}\right) = 1$ On further simplification we find that $16\cdot 6^n - 15\cdot 9^m =1 \phantom{cc} {\color{#D61F06} 16\cdot (3^n)2^n - 5\cdot 3^{2m+1} =1 }$ As Sir @Pi Han Goh makes an observation that left hand side is divisible by 3 however, the right hand side isn't divisible by 3 . Hence, there is no solution for $m$ and $n$ .

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Another approach $6^n = \dfrac{1+15\cdot 9^m}{16}$ On the right hand side we note that $\ 1+15\cdot 9^m \equiv \begin{cases} 0 \mod(16) \phantom{ccccc} \text{if m is even} \\ \pm 8\mod(16)\phantom{cccc}\text{if m >1 is odd} \end{cases}$

In odd case we are not going to deal since it doesn't yield any positive integer and we are left with even case. Further we note that $6^n \equiv 0\mod(6)$ however, RHS of the above equation is found to be $k = \dfrac{1 +15\cdot 9^m}{16} \equiv \begin{cases} -4\mod(6)\quad \text{ if m = }2\times {\text{odd}} \\ -1\mod(6) \quad \text{if m = }4\times Z^+\end{cases}$ This shows that it is not possible to find positive integers $m$ and $n$ such that desired equations can be true .