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Motivation: Brocard’s problem $n!+1$ being a perfect square

Observations: Given a power Diophantine equation of $k$ variables with a “general solution” (provides infinite integer solutions) to the equation which makes the equation true for any integer. Following can be said-

1. The “general solution” (provides infinite integer solutions) is a rational(not necessary), algebraic expression(can have multiple variable), .
2. The general solution can be expressed using $1$ variable, so the given equation can be expressed using function of $1$ variable and finally, the derivative of the both side will be equal.

In other words, given an, exponential Diophantine equation in $k$ variables with infinitely many solutions, there is some family of uni-variate,closed-form formulas $R_1(n),\ldots,R_k(n)$ (not all constant) using the formal variable $n$, the field operations (+, -, $\times$, /), the integers, and exponentiation, such that $R_1(n),\ldots,R_k(n)$ is a solution for all integers $n$.



Example: It is known,$x=a^2-b^2;y=2ab; z=a^2+b^2$ are “general solutions”to a power Diophantine equation $x^2+y^2=z^2$, here $k=2$,for any integer $a,b$ the equation holds true. ,$x=a^2-b^2;y=2ab; z=a^2+b^2$ are rational, algebraic expression with $2$ variables $a,b$. It can written that,$x=f_1(a,b),y=f_2(a,b),z=f_3(a,b)$ Keeping $b$ constant,$x=g_1(a),y=g_2(a),z=g_3(a)$ , thus, general solution is expressed using $1$ variable, namely, $a$, where $b$ is fixed/ constant. So, the equation becomes,$g_1(a)^2+g_2(a)^2=g_3(a)^2$, and the derivative of the both side is same with respect to $a$.

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).

Question: This is a verification post (as tagged below). Are those Observations right? What are the flaws?

Request: Please do not provide another solution (elliptic curve, divisibility). Answer should be directly regarding the post.

Remarks: 1. 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 $a=1$.
2. The equality of derivative is a "necessary" condition, not "necessary and sufficient" condition, so, it will not have any impact on FLT for $n>2$.

3.Dont consider different sort of function like, the trivial equation $x−y=0$. you can find the following general solution: Take $g_1(a)=a $and take $g_2(a)=acos(2πa)$. Then , $g_1(a)−g_2(a)=0$ for each $a∈Za\in \mathbb{Z}$. But the derivatives are not equal.(please help to define this!)

Please let me know if anything is unclear / undefined.

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Michael
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    Consider the equation $$n!+1=k(n+1).$$ Wilson's theorem says that $k$ is an integer if and only if $n+1$ is a prime number. Therefore this equation has infinitely many solutions. Yet it has no general solution of the form you talk about (there is the famous polynomial in zillion variables with the property that whenever its value is positive that value is a prime, but this cannot be done with a single variable). – Jyrki Lahtonen Jun 03 '15 at 18:53
  • Sir, please forgive me if this is stupid question,1. Should I consider wilsons thm case as a power diophantine equation 2. if there is no general solution at the moment, it might be discovered later. is it proved that there could be no general solution to that eq? 3. Sir, would you please consider without the motivation and comment what I described. Thanks. – Michael Jun 03 '15 at 19:04
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    I would say that factorial definitely cannot appear in a power equation (do you mean a polynomial equation??). I only produced that example because I knew you were interested in Brocard, and I wanted to give an example of an equation that has infinitely many integer solutions, but those solutions do not come from a simple formula. This was largely an attempt to get you to make your question more precise. – Jyrki Lahtonen Jun 03 '15 at 19:08
  • the problem you wrote is related to prime number directly, so if one finds the formula of constructing such prime, there is a very good chance that one would find a general solution which would be highly irregular as primes are irregular, it can be said without any doubt that the formula to find prime is extremely hard to construct , but it is not proved that such a formula is impossible to find, right?(polynomial formula is impossible) anyway , there could be result, i dont know of, so, is it proved that there could be no general solution to that eq? – Michael Jun 03 '15 at 19:28
  • you wrote "there is the famous polynomial in zillion variables with the property that whenever its value is positive that value is a prime, but this cannot be done with a single variable)", please provide information/link regarding the subject. – Michael Jun 03 '15 at 19:41
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    See here or here. – Jyrki Lahtonen Jun 03 '15 at 20:49
  • @ Jyrki Lahtonen,Sir, my limited, little knowledge tells me borcards problem is different than the problem $n!+1=(n+1)k$ due to prime number formula. Still It is not proved that it is impossible to construct a prime number formula(may be the way I am using might do it).I can use this post's concept in Brocards problem (if it is correct). In the post, i wrote about rational expression, actually it is not necessary to be rational. It would be help full if you comment on observation as i plan to use it in another post. Thanks for your precious time. – Michael Jun 04 '15 at 12:17

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