Solving algebraic equations involving factorials - without trial and error.

Question

The question is defined as follows:

$$\frac{(x!)^3}{x}-1=3455$$

I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression as seen below:

$$x!=3456^{1/3}\cdot x^{1/3}$$

$$x(x-1)!=x\cdot 3456^{1/3}\cdot x^{-2/3}$$

$$3456^{1/3}/(x-1)!=x^{2/3}$$

I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.

So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?

Answer 1

We have $3456=2^7\cdot 3^3$. So, you have

$$(x!)^3=3456x=2^7\cdot 3^3\cdot x.$$

In particular, no prime $>3$ can divide $x!$, and so $x\leq 4$. You also need $2^7 \cdot 3^3 \cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.

Answer 2

"Without trial and error" seems excessive, as the function is very quickly growing.

Grossly, you can use

$$x!^2\le \frac{x!^3}{x}\le x!^3$$ which gives you the range

$$x!\in(15,59)$$

and only $4!$ can do.

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But my first reaction would be to try $3$ to $5$ without any preliminary effort.

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It is worth to notice that Stirling is not helpful here.

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For larger values, I would consider the bracketing between $\frac{\log(m+1)}{3\log\log(m+1)}$ and $\frac{\log(m+1)}{2\log\log(m+1)}$ and work by dichotomy.

Answer 3

Let $\,a_n=\dfrac{\left(n!\right)^3}{n}\,$ then $\,a_{n+1}=n(n+1)^2a_n\,$. It's quite obvious that $\,a_n\,$ grows very fast, and the first few terms are $\,a_1=1, a_2=4, a_3 = 72, a_4=3456\,$. The latter gives the solution $\,x=4\,$.