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secondprime
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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##.
$$x^{2}-1=n!$$
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##.
