Condition for Power Diophantine Equation

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1. Jun 4, 2015

secondprime

**Observations:**
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.

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.

2. Jun 4, 2015

Staff: Mentor

It's not clear what you're trying to say here, possibly due to language differences. By a power Diophantine equation of k variables, I assume you're talking about an equation of this form:
$x_1^n + x_2^n + \dots + x_k^n = y^n$, where $x_1, x_2, \dots, x_k$ are integers.
Your use of the word "and" above is throwing me off. Are you saying that we have a Diophantine equation like the one I show, and that it has a general solution? If so, I don't understand. This is like saying, "If x = 2, then."
My questions is, "Then what?"

Or are you saying that for such an equation there exists a general solution?
What are $g_1, g_2, g_3$? I can see that you explain what they are below, but you should explain what they are right when these terms are introduced.
Also, this is quite a leap, from saying that the general solution can be expressed as a function of one variable, and that the derivatives of both sides of some equation are equal. Why?
How do $g_1, g_2, g_3$ "move to another $g_1, g_2, g_3$"? What does this mean?
If you hold b constant, what guarantee is there that x, y, and z will be integers?
"and if the derivative of the both side is the same with respect to a"
Then what?
What are the derivatives of the two sides?
If you have two functions that have the same derivative, can you conclude that the two functions are equal? (Answer: no.)
FLT = ?

3. Jun 5, 2015

secondprime

There has been an arrangement problem, please check this post, let me know if any edit work is needed.
Below 2 facts are given -
1.A power Diophantine equation of $k$ variables.
2. there exists a “general solution” (provides infinite integer solutions) to the equation which makes the equation true for any integer.

Observations are

1. The “general solution” (provides infinite integer solutions) is a rational/irrational, algebraic expression(can have multiple variable), since transcendental function/expression will not provide 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.

**Example:**
Consider a case where $k=2$.
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 be 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, 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:** 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(Fermat's Last Theorem) for $n>2$.

power might not be same.
it is Pythagorean triplet, true for any integers $a,b,c$.

I agree but I have not written anything like that, I am equating both sides.

4. Jun 5, 2015

Staff: Mentor

In my opinion, it is not an arrangement problem - it is a problem in how you are explaining what you are doing.
Please define what you mean by "power Diophantine equation of k variables." I thought I understood what you meant, but my guess was apparently wrong. The last time I took a class in number theory was in 1976, so please refresh my memory on the terms you're using.
Without knowing what the equation is, I don't know how to confirm that the equation is true for any integer. And besides, you said the equation has k variables.
I asked several questions in my previous post, most of which you either ignored or gave "answers" that didn't really answer my question.

Now you seem to be talking about $x^2 + y^2 = z^2$. How does what you're saying apply to the more general case, where there are k varables?

Then what is your point in equating the two derivatives?

5. Jun 5, 2015

secondprime

Sir, I am sorry for you inconvenience, nothing has changed, I will try my best to clarify.
Sir, these are basic assumption. you assume these facts to be true and consider 2 observations under these 2 assumption.

I do not dare to ignore your question. I thought re-posting would clear. (FLT = fermats last theorem.)

for example, $x^2+k=y^3+ z^5$ , here k=3 variables,namely x,y,z, each has different power.

The general solution is bases on any integer, thus it can be applied to general case.

it will be used later in another post, would verify under this post's circumstances?

6. Jun 5, 2015

Staff: Mentor

It is not a good idea to have k represent the number of variables and also to appear in the equation for what appears to be a different purpose.
What does this mean?

As I understand things, there are two assumptions:
1. There is an equation, such as $x^2+k=y^3+ z^5$.
2. There exists a “general solution” (provides infinite integer solutions) to the equation which makes the equation true for any integer.
Would this be $x = f_1(a, b), y = f_2(a, b), z = f_3(a, b)$?
For the specific equation x2 + y2 = z2, you showed earlier that if x = a2 - b2, y = 2ab, and z = a2 - b2 are substituted into x2 + y2 = z2, this equation is identically true for any choice of a and b.

Fine, I agree with that.
For your other example, $x^2+k=y^3+ z^5$, how can I be sure that assumption 2 above is a reasonable one to make? Can you show me the three functions f1, f2, and f3? You can make any assumption you like, but if you start with a false assumption, then logically, any conclusion you reach is true, but meaningless.

I would like to see at least a brief description of why this is important to do, in this thread, not in some other thread.

7. Jun 6, 2015

secondprime

it can be multi-variable, more than 1 variable may be required for general solution.
as answered above, you can have more than 1 variable, keeping one variable and other constant, you can be sure of what I said.
it is regarding brocards problem which you closed.

8. Jun 6, 2015

Staff: Mentor

My question was regarding the general solution of the equation you gave as an example, $x^2+k=y^3+ z^5$. What would the general solution that you talked about look like?
No, you did not answer my question, otherwise I would not have asked it here. You are starting with two assumptions that I listed in post #6. Before we can discuss your observations, I need to be assured that your second assumption is a reasonable one to make. Please show me how a general solution to $x^2+k=y^3+ z^5$ could be attained.
Which is the likely fate of this thread if you are unable to answer the questions I'm asking.

9. Jun 6, 2015

secondprime

it was an example of power Diophantine equation , The human race don't know general solution of every power Diophantine equation. I already gave an example of the idea, why you need this particular equation's example?
looks like a threat to me !! is it really necessary ?

Last edited: Jun 6, 2015
10. Jun 6, 2015

Staff: Mentor

I'm not asking for the general solution of every power Diophantine equation -- just the one that you gave - $x^2 + k = y^3 + z^5$.
Apparently, yes. The policy of this forum is to discuss mainstream concepts in science (and mathematics), not to promote personal theories, which appears to be what you are doing in this thread and the earlier one about Brocard's problem. I was willing to see where you were going with this one, despite the forum policy, but from what I see in this thread, your proof rests on certain assumptions that I have serious problems with.

11. Jun 6, 2015

micromass

I think he only wants to discuss those equations which do have a general solution. Whether many equations have such solutions seems irrelevant to him, he just ignores the ones without general solution. So it could be that the class of equations he is considering is very tiny. So I think we should probably let this point go.

12. Jun 6, 2015

micromass

I have a problem with this. As of now, the natural of the functions $g_k$ implies that you only look at those $a$ for which $g_k(a)$ is an integer. But then your equation $g_1(a)^2 + g_2(a)^2 = g_3(a)^2$ is not satisfied for all $a$, only for those for which $g_k(a)$ is an integer. So if we don't have a full equality on $\mathbb{R}$, why would the derivatives be the same?

13. Jun 6, 2015

secondprime

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).
$g_k$ is continuous for all a.

consider each function separately, they are continuous interdependently and differentiable since each of them is a defined as a closed form formula/ algebraic expression.

14. Jun 6, 2015

micromass

OK, so you still don't understand what I mean. Consider the trivial equation $x-y=0$. Then I can find the following general solution: Take $g_1(a) = a$ and take $g_2(a) = a\cos(2\pi a)$. Then we indeed have that $g_1(a) - g_2(a) = 0$ for each $a\in \mathbb{Z}$. But the derivatives are not equal.

15. Jun 6, 2015

secondprime

Nice example !! I understand now.
but if I am not wrong two function of your example seem different to me, I think $cosx, sinx$ goes to transcendent class( Lindemann-Weierstrass) , I did not have this kind of situation in my mind though I wrote "could be transcendental,irrational function/expression if provides integer solution"
for the time being would you please consider that the solution would not have different expression like your example?
are you suggesting that there are solution which can have different function like your example?

16. Jun 6, 2015

micromass

Well, then you need to specify a good class of general solutions $g_k$, for example polynomials. And then you need to prove that for this class, the derivatives do equal. But I hope you see that a proof is needed, since it is not automatically.

17. Jun 6, 2015

secondprime

@micromass, dont you think its highly unlikely that general solution would have different kind/class of expression??

18. Jun 6, 2015

micromass

That's not a proof.

19. Jun 6, 2015

secondprime

understand, consider the equation class considered above does not have different kind/class of function/ algebraic expression as a solution.

20. Jun 6, 2015

certainly

21. Jun 6, 2015

secondprime

Do you have any specific question @ certainly?

22. Jun 6, 2015

^ This.