# Brocard's problem [edited]

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1. Apr 17, 2015

### secondprime

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's_problem.
According to Brocard's problem
$$x^{2}-1=n!=5!*(5+1)(5+2)...(5+s)$$
(x,n) is the solution tuple of the problem. If there are infinite $x, n$for which the above equation is true, then for each of $x$, there is exactly one $s$. It is a "one-to-one" relation. Therefore a "well defined","single valued" function f(x) exists which maps $x$ to $s$(an element of set of all s), so $s=f(x)$

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 $\mathcal{O}(5^{s})$ as all terms which grow$\leq 5^{s}$

Since,$s=f(x)$, so,$5^{s}=5^{f(x)}$ but,

$\frac{d}{dx}x^{2} <\frac{d}{dx} \mathcal{O}(5^{f(x)})$(after certain x). Therefore,

$\frac{d}{dx}x^{2} <\frac{d}{dx} \mathcal{O}(5^{f(x)})<\frac{d}{dx}5!*(5+1)(5+2)...(5+s)$
After certain x, it is not possible to hold $x^{2}-1=n!=5!*(5+1)(5+2)...(5+s)$.
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 $5^m$(because $f(x)$ has to be integer) and the equation becomes,$5^{2m}-1=k *\mathcal{O}(5^{r})=n!$ which is not possible, since $5\nmid n!+1$ when $n>5$.

2. Apr 17, 2015