### 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.

$$x^{2}-1=n!$$

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.