Brocard's problem is a problem in mathematics that asks to find integer values of n for which(adsbygoogle = window.adsbygoogle || []).push({});

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

http://en.wikipedia.org/wiki/Brocard's_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## ,

$$\frac{d}{dx}x^{2} <<\frac{d}{dx} \mathcal{O}(5^{f(x)})$$

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## but, ##{2 \over \log 5}\log x## is only once integer (when, x=5), if Brocard's problem has infinte solution then f(x) must have infinite integer values. So, ##f(x) \neq {2 \over \log 5}\log x##.

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.

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# An argument for "Brocard's problem has finite solution"

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