- #1

secondprime

- 49

- 0

**Given a power Diophantine equation of ##k## variables and there exists a “general solution” (provides infinite integer solutions) to the equation which makes the equation true for any integer.**

**Observations:**

**Observations:**

1. The

__“general solution”__(provides infinite integer solutions) is

__an algebraic expression__(

*can have multiple variable,rational, could be transcendental,irrational function/expression if provides integer solution*) .

2. The general solution can be expressed using ##1## variable, so the given equation can be expressed using function of ##1## variable and thus the both side of the derivative will be equal. 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).

****Example:****

##k=2## case.

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##,for any integer ##a,b## these solutions works.

,##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 if the derivative of the both side is same with respect to ##a##

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

****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##.

** future edit is possible to clarify.