<s>[INST] Can 720! be written as the difference of two positive integer powers of 3? Does the equation:
$$3^x-3^y=720!$$
have any positive integer solution?
 [/INST] As alex.jordan writes, $3^x-3^y$ factors as $3^y(3^{x-y}-1)$, so $y$ must be the number of factors of $3$ in $720!$.
I don't actually need to count the number of $3$s in $720!$, so let's just define the notation $720!_3$ for "$720!$ with all of the powers of $3$ divided out". This must yield the other factor $3^{x-y}-1$, so we need to investigate whether $720!_3$ is one less than a power of $3$. To do this we will compute it modulo $3$.
In general we have that
$$ (3k)!_3 \equiv (1\cdot 2)^k \cdot k!_3 \equiv (-1)^k k!_3 \pmod 3 $$
And therefore
$$ \begin{align} 720!_3
\equiv 240!_3 
&\equiv 80!_3 \equiv 80\cdot 79 \cdot 78!_3 \equiv - 78!_3 \\
&\equiv -26!_3 \equiv -26\cdot 25 \cdot 24!_3 \equiv 24!_3 \\
&\equiv 8!_3 \equiv 8\cdot 7\cdot 6!_3 \equiv -6!_3 \\
&\equiv -2!_3 \equiv -2 \equiv 1 &\pmod 3
\end{align}$$
which is not one less than a multiple of 3, so certainly not one less than a power of $3$.
So the answer is no.
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