# Solve this equation: $3^{3x} - 3^x = (3x)!$

I have this equation:

$$3^{3x} - 3^x = (3x)!$$

We have to solve for $$x$$ integer. I did try to attempt but to no avail. I can't manipulate any side of this equation. I took common $$3^x$$ in the LHS of the equation and got a product: $$(3^x) (3^{2x}-1)$$ but I have no idea what to do in the RHS of the equation (which is a factorial). It looks like the answer is $$x=2$$ but I want to solve it algebraically.

Any hints/solution would be greatly appreciated.

• Are you looking for an integer solution? otherwise the expression $(3x)!$ makes no sense (unless you introduce the gamma function). Jul 14, 2021 at 10:31
• @FeedbackLooper Yes, I am looking for an integer solution. This equation equates for x=2. I want to solve it algebraically. Jul 14, 2021 at 10:33
• Actually, I suppose one could a priori only demand that $3x$ be an integer. But then $3^x$ is forced to be an integer, which in turn forces $x$ to be an integer. Jul 14, 2021 at 10:35
• @user123 Added to the post itself for you, next time please include such details in the post itself, not just a comment. Anyway one strategy could be to try to show that right hand side is greater than the left hand side for $x \geq 3$ (induction?).
– Sil
Jul 14, 2021 at 10:36

The main idea is that $$n!>a^n$$ for $$n$$ sufficiently large, so there is only a finite number of values to check.

In this problem, a simple mathematical induction shows that $$n!>3^n$$ for every $$n\ge 7$$. Therefore, for $$x\ge 3$$, we have $$(3x)! > 3^{3x} > 3^{3x} - 3^x$$.

Now, the equality is satisfied for $$x=2$$ ($$3^6-3^2=720=6!$$) but not for $$0$$ ($$3^0-3^0=0\neq 0!$$) or $$1$$ ($$3^3-3^1=24\neq 3!$$)

• (+1) Excellent answer! I particularly like its simplicity, and how you don't need an advanced level of knowledge to understand it. Jul 14, 2021 at 12:56

Here is a way by comparing exponents of $$3$$ on both sides for $$x \geq 3$$.

Notice that RHS is divisible by $$(3x)(3x-3)(3x-6)\cdots(3)=3^x(x(x-1)\cdots 1)=3^xx!.$$ Because $$x \geq 3$$, this means $$3 \mid x!$$ and in turn $$3^{x+1}$$ divides RHS. But that means it must divide LHS as well, i.e. $$3^{x+1} \mid 3^{3x} - 3^x = 3^x(3^{2x} - 1)$$ (the equality you have found). But this means $$3 \mid 3^{2x}-1$$, impossible.

A factorial can be written as a product of three subsequent integers only in very few cases, in particular: \begin{aligned}1\times 2\times 3&=3!\\2\times 3 \times 4=24&=4!\\4\times 5\times 6=120&=5! \end{aligned} or $$8\times 9\times 10=720=6!$$ LHS of the given equation is $$3^{3x} - 3^x = (3^x-1)(3^x)(3^x+1),$$ which is a product of subsequent integers where the mid-term is a power of $$3.$$ Comparing with the listed possibilities, only the last one satisfies. Here $$3^x=9$$ or $$x=2.$$

• Nice answer. How do you justify that a product of three consecutive integers is a factorial only in the few cases you mentioned? Jul 15, 2021 at 0:51
• This answer indicates it might be an open problem...
– Sil
Jul 15, 2021 at 8:25
• @Sil thank you for the link, I could not find it (or a similar) yesterday. I find the question challenging and the solutions nice. Jul 15, 2021 at 12:50
• @Taladris My first idea was thinking of the Gamma function. Then I noticed that OP did not accept your excellent solution. Thus I decided to write my solution as simple as possible, maybe for high-school level. I agree there is a gasp :) One possible generalization of the given problem is to solve in integers $n!=k!m!,$ because the equation can be written as $(3^x+1)!=(3x)!(3^x-2)!$ For this, see the link, which could complete my answer. Jul 15, 2021 at 13:01