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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.

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    $\begingroup$ Are you looking for an integer solution? otherwise the expression $(3x)!$ makes no sense (unless you introduce the gamma function). $\endgroup$
    – FeedbackLooper
    Jul 14, 2021 at 10:31
  • $\begingroup$ @FeedbackLooper Yes, I am looking for an integer solution. This equation equates for x=2. I want to solve it algebraically. $\endgroup$
    – user123
    Jul 14, 2021 at 10:33
  • $\begingroup$ 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. $\endgroup$
    – Aryaman Maithani
    Jul 14, 2021 at 10:35
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    $\begingroup$ @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?). $\endgroup$
    – Sil
    Jul 14, 2021 at 10:36

3 Answers 3

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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 $1$ ($3^3-3^1=24\neq 3!$)

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    $\begingroup$ (+1) Excellent answer! I particularly like its simplicity, and how you don't need an advanced level of knowledge to understand it. $\endgroup$
    – A-Level Student
    Jul 14, 2021 at 12:56
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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.

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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 $x=2.$

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    $\begingroup$ Nice answer. How do you justify that a product of three consecutive integers is a factorial only in the few cases you mentioned? $\endgroup$
    – Taladris
    Jul 15, 2021 at 0:51
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    $\begingroup$ This answer indicates it might be an open problem... $\endgroup$
    – Sil
    Jul 15, 2021 at 8:25
  • $\begingroup$ @Sil thank you for the link, I could not find it (or a similar) yesterday. I find the question challenging and the solutions nice. $\endgroup$
    – user376343
    Jul 15, 2021 at 12:50
  • $\begingroup$ @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. $\endgroup$
    – user376343
    Jul 15, 2021 at 13:01

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