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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%27s_problem. According to Brocard's problem $$x^{2}-1=n!=5!*(5+1)(5+2)...(5+s)$$ here,$(5+1)(5+2)...(5+s)=\mathcal{O}(5^{s}),5!=k$. So, $$x^{2}-1=k *\mathcal{O}(5^{s})$$ Here, $\mathcal{O}$ is Big O notation. For every rational number $x$ , there is a rational $s$( since $k$ is a constant, if $x$ is increased, $s$ has to be increased to balance the equation). It is a "one-to-one" relation, so there exists a "well-defined" function $f(x)$, so $s=f(x)$ $$x^{2}-1=k *\mathcal{O}(5^{f(x)})$$

Claim: Above equation can not have infinite solution, because 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 infinite problem then $f(x)$ must have infinite integer values. So, $f(x) \neq {2 \over \log 5}\log x$.

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

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


Remarks:
1.you can use other integer value instead of 5.
2. there are terms(factorial), in the expansion of $(5+1)(5+2)...(5+s)$, with bigger growth than $5^{s}$ which is used for simplicity.

edit 1: It is known,$x=a^2-b^2;y=2ab; z=a^2+b^2$ are solutions to$x^2+y^2=z^2$,for any integer $a,b$ these solutions works.It can written that,$x=f_1(a,b),y=f_2(a,b),z=f_3(a,b)$ Keeping $b$ constant,$x=g_1(a),y=g_2(a),z=g_3(a)$.

So, for each integer $a$, an integer solution of the equation can be found and as $a$ increases $g_1,g_2,g_3$ increases, because when these $g_1,g_2,g_3$ are plotted on 2D, it can be noticed that from one integer point to another bigger integer point,each of these $g_1,g_2,g_3$ moves continuously to another $g_1,g_2,g_3$ as $g_1,g_2,g_3$ are algebraic, rational expression (otherwise not possible to provide integer solution with transcendental function expression). Same idea is used above.

In general, $g_1,g_2,g_3$ might have points where these function are not differentiable but those points are finite, so the argument can be carried on after those points. For example, if $g(a)= \frac{5a}{a-1}$ then $g$ is continuous after $a=1$.

Why the above argument is not correct? what are the flaws?

Michael
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    I think that you can not say that $x^2 -1 = k \cdot \mathcal O(5^r)$, because as $r$ is getting bigger, there number of factors in factorial is getting bigger too. –  Apr 12 '15 at 17:10
  • $\mathcal{O}(5^{r})$ represents the biggest term of a polynomial $(5+1)(5+2)...(5+r)$. If you consider the equation, r increses as much as x increses. Your concern of r regarding n! is alright , but you need to keep in mind that this r must satisfie another "restriction" due to the equation. – Michael Apr 12 '15 at 17:13
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    But what you are trying to say is that $n! = \mathcal O(5^{n-5}) = \mathcal O(5^n)$ which is false. –  Apr 12 '15 at 17:14
  • plesae read the previous comment. $\mathcal{O}(5^{r})$ is as much as it need to be a part of n!, I did not give an explicit formula for r, that is not my concern, whatever r is (the bigger growth it has , the merrier), it must have chage rate of right hand side. I never mean ,$ n! = \mathcal{O}(5^{r})$ – Michael Apr 12 '15 at 17:18
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    $n! = x^2-1 = k \cdot \mathcal O(5^r)$, right? But $r = n-5$, so $\dots = k \cdot \mathcal O(5^r) = 120 \mathcal O (5^r) = \mathcal O (5^r \cdot 120)$ and this is equal to $\mathcal O(5^{n-5})$ which is $\mathcal O(5^n)$. Believe me, if this problem had such a simple solution, somebody would already find it. –  Apr 12 '15 at 17:23
  • Let us continue this discussion in chat. – Michael Apr 12 '15 at 17:30
  • What's $f(3)$ ? – mercio Apr 12 '15 at 17:51
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    You don't explain why the rate of change of (something the size of) $5^{f(x)}$ must be bigger than that of $x^2$ - indeed, as you say yourself, there are choices of $f(x)$ that make the two sides identical. You can't simply say that '$5^{f(x)}$ is exponential and therefore has higher growth rate' since $f()$ is an arbitrary function. Does $5^{\log_5(\log_5(x))}$ have faster growth rate than $x^2$? – Steven Stadnicki Apr 14 '15 at 15:09
  • It is not essential to be bigger, it might be $< \frac{d}{dx}x^{2}$, either way, it violates the equation. Choices of $f(x)$ that make the two sides identical, are limted, the one I described above is worthless. It would be helpful if you suggest how I can find such functions that make the change rate of two sides identical. – Michael Apr 14 '15 at 17:33
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    I agree with Leon. I cannot make sense of your big O. The product $(5+1)(5+2)\cdots(5+r)$ is not $\mathcal{O}(5^r)$. It is way bigger. – Jyrki Lahtonen Jun 05 '15 at 05:40
  • I have edited , there is no $r$ anymore except in the [ ] , it remained there as a mistake. consider $s$ instead $r$. If it is bigger then it will not have the same change rate/ derivative as right hand side, that should go in favor of my argument. – Michael Jun 05 '15 at 05:57
  • Replacing $r$ with $s$ does not make false fact true. – Jyrki Lahtonen Jun 05 '15 at 06:03
  • Agree, just informing the change. – Michael Jun 05 '15 at 06:04
  • @JyrkiLahtonen, Sir, why did you delete your answer? – Michael Jun 05 '15 at 06:06
  • I no longer think it addresses your question. – Jyrki Lahtonen Jun 05 '15 at 06:08
  • @JyrkiLahtonen,Sir, there are 2 issues that need to be addressed.
    1. Is it possible to have such terms for which growth of that terms are less than $2x$ and at the same time , to have terms for which growth of that terms are more than $2x$ so that over all change rate is $2x$.
    2. Why there are solution more than one.

    Answer to 1st question is negative because there is an factorial term , that is sure to be integer and must have bigger growth than $2x$ for sufficient large $x$.

     – Michael Jun 05 '15 at 06:29
  • Answer to 2nd question is “curve intersection”. Algebraic curve intersects once, comes back and intersects again with another algebraic curve, similar phenomenon is also happening here. I think,atleast, It can be proven that there is no “general solution” or “algebraic expression as a general solution” for brocards problem(atleast,if this is true -http://math.stackexchange.com/questions/1310966/necessary-condition-for-power-diophantine-equation). – Michael Jun 05 '15 at 06:29
  • It is just unclear to me what you mean by all those term. Points on an algebraic curve can be parametrized by rational functions if and only if that curve has genus zero. On genus one curves (elliptic curves) we know that there can be infinitely many rational points but only finitely many integers points (IIRC), Faltings' theorem handles the higher genus curves. But Brocard is not related to an algebraic curve at all, because it has that factorial, or Gamma-function if you prefer. – Jyrki Lahtonen Jun 05 '15 at 06:40
  • If you are really only claiming that there cannot exist functions $f$ and $g$ of some class (polynomial, rational, algebraic, exponential,...) such that $x=f(t)$, $n=g(t)$ is a solution to Brocard for all integers $t$, then you should formulate the question that way. As it is written all that talk about big-O is difficult to filter out, because the misunderstandings there need to be addressed first. – Jyrki Lahtonen Jun 05 '15 at 06:44
  • terms are obtained from expansion of $(5+1)(5+2)...(5+s)$, e.g, $5^s$ is a term form that expansion , as i would like to write $s=f(x)$, it becomes $5^{f(x)}$,. The expansion has factorial terms depending on $s$, thus on $f(x)$, so there is a term in the expansion, say, $g(x)!$ which will have bigger growth (where $g(x) $ is integer). I am not relating algebraic curve directly(at the moment), I am just giving an example of similar phenomenon, probably I can use gamma function to show the bigger growth of $g(x) $. I will consider you advice regarding the confusion. Thanks for the advice,Sir . – Michael Jun 05 '15 at 06:53
  • I don't think you can use derivatives; as far as the problem is concerned the points that $n!$ can take are discrete. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Feb 10 '19 at 10:49

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