Page 1 of 1

### Brocard's problem has finite solution(?)

Posted: Tue May 05, 2015 12:08 pm
Brocard's problem is a problem in mathematics that asks to find integer values of n for which
$$x^{2}-1=n!$$
http://en.wikipedia.org/wiki/Brocard%27s_problem.
According to Brocard's problem
$$x^{2}-1=n!=5!*(5+1)(5+2)...(5+s)$$
here,$(5+1)(5+2)...(5+s)=\mathcal{O}(5^{r}),5!=k$. So,
$$x^{2}-1=k *\mathcal{O}(5^{r})$$
Here, $\mathcal{O}$ is Big O notation. For every rational number $x$ , there is a rational $r$( since $k$ is a constant, if $x$ is increased, $r$ has to be increased to balace the equation). It is a "one-to-one"
relation, so there exists a "well-defined" function $f(x)$, so $r=f(x)$
$$x^{2}-1=k *\mathcal{O}(5^{f(x)})$$

**Claim:** Above equation can not have infinite solution, becuase change rate of $\mathcal{O}(5^{f(x)})$is much bigger than $x^{2}-1$ ,

$$\frac{d}{dx}x^{2} <<\frac{d}{dx} \mathcal{O}(5^{f(x)})$$ or
$$\frac{d}{dx}x^{2} >> \frac{d}{dx} \mathcal{O}(5^{f(x)})$$ in gennral,
$$\frac{d}{dx}x^{2} \neq \frac{d}{dx} \mathcal{O}(5^{f(x)})$$
[

1.if $f(x) = {2 \over \log 5}\log x$ then, $5^{f(x)} = \mathrm e^{f(x) \log 5} = \mathrm e^{2 \log x} = x^2$ so, $x$ has to be $5^m$ and the equation becomes,$$5^{2m}-1=n!$$

2.if $5!*(5+1)(5+2)...(5+s)$ is expanded, there are terms which grow $\leq 5^{s}$ and terms which grow $\geq 5^{s}$.
Consider $\mathcal{O}(5^{r})$ as all terms which grow$\leq 5^{s}$ in above case.

3.There exists factorial term in the expansion(e.g. $s!$),in this case again the function $f(x)$ can be associated and shown that the growth of left-hand side is bigger than the right-hand side in terms of x which is not acceptable for an equation of infinite solutions .

]

So, after certain value of $x$, the equation will not hold.

**Why the above argument is not correct? what are the flwas?**
** you can use other integer value instead of 5.

### Re: Brocard's problem has finite solution(?)

Posted: Mon Jul 05, 2021 7:13 am
The claim that $(5+1)(5+2)(5+3)...(5+s)=O(5^{k})$ is false. Big O notation requires the ratio between the two functions have a maximum value when given the same input. Since the first function is multiplied by a number bigger than five, while the other is multiplied by 5, this means that $\lim_{x \to +\infty} \frac{f(x)}{5^{k}}$ diverges, so the usage of the big O notation is wrong. Note that changing 5 for something else wouldn't fix the problem that factorials eventually outgrow all exponentials.

The chaim that the change rate of $O(5^{f(x)})$ is much bigger than $x^{2}−1$ depends on the growth of $f(x)$. For example, if $f(x)$ was 0 for all x, then $O(5^{f(x)})$ would be constant, it wouldn't grow at all. In fact, because the sides of the equation $x^{2}−1=k∗O(5^{f(x)})$ have to be equal, f(x) would be a slow growing function in order to conpensate for the exponential. In particular, $f(x)=\log_{5}{x^{2}}$ would simplify that statement to $x^{2}-1=O(x^{2})$, satisfying the equality while also being a slow growing function.