Solve the Pythagorean Tripple 3^x+4^x=5^x

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around the equation 3^x + 4^x = 5^x, exploring methods to solve it and the implications of Fermat's Last Theorem. Participants consider both integer and non-integer solutions, as well as the algebraic techniques that might be applied to find solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants recognize the equation as a Pythagorean triple but seek methods to solve it without prior knowledge of the solution.
  • One participant asserts that Fermat's Last Theorem implies the only integer solutions are x = 0, 1, and 2, with x = 2 being valid.
  • Another participant questions how to determine non-integer solutions and suggests exploring other forms of the equation, such as 4^x + 10^x = 13.4534543^x.
  • Some participants propose using logarithms or integration as potential methods for solving the equation.
  • One participant mentions the Intermediate Value Theorem in relation to finding solutions for a modified equation, indicating that solutions may exist but may not be solvable by standard methods.

Areas of Agreement / Disagreement

Participants express differing views on the existence of non-integer solutions and the applicability of various mathematical methods. There is no consensus on a definitive method for solving the original equation or the modified examples presented.

Contextual Notes

Participants note limitations regarding the assumptions of integer solutions and the challenges of applying standard mathematical techniques to find solutions for the equations discussed.

Diffy
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The problem is 3^x + 4^x = 5^x. I recognize that this is a pythagorean triple but I am curious as to how you could solve this without just knowing that it is a pythagorean triple.
 
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FLT states that the only way all of those solutions could be integers is if x=2.
 
I am sure Fermats Theorem says what you say it says, however I am not sure that answer my question. I am trying to see if there is a way to solve the equation I have posted without just knowing the answer is 2.
 
Solve what equation? 3^2+4^2 = 5^2 is obviously true. Are you looking for a general way to generate a three positive integers (a,b,c) such that a^2+b^2=c^2?
 
I guess I am looking for an algebraic way to solve 3^x + 4^x = 5^x for x. (sorry if I used the term algebraic incorrectly). So what I mean is take the natural log of both sides, then do this etc. Or perhaps integrate both sides, then do something else etc.

Is it possible?
 
Diffy said:
I guess I am looking for an algebraic way to solve 3^x + 4^x = 5^x for x. (sorry if I used the term algebraic incorrectly). So what I mean is take the natural log of both sides, then do this etc. Or perhaps integrate both sides, then do something else etc.

Is it possible?

No. It's called Fermat's last theorem.

Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation [tex]a^x + b^x = c^x[/tex] for any integer value of x greater than two.

So, assuming x is an integer, the only possible answers are 0, 1, and 2. Obviously, x = 2 works and the rest fail.
 
Fermat-Wiles theorem. Give the guy credit who proved it along with the one who conjectured it.
 
Ok. Fine. Say I didn't use 3,4,and 5 as a, b, and c. HOw do we know there are no non integer solutions.

I'm making this up off the top of my head, so say we had 4^x + 10^x = 13.4534543^x How would you solve for x?

0, 1 and 2 don't work obviously, if there were any other solutions for x, how would we solve it, if at all...
 
That is possibly a very good question...

Could we take the log of it?Maybe. Would the answer then make sense?... Maybe. The question
is... do we need an integer solution of x?
 
  • #10
[tex] \begin{array}{1}<br /> f(x) = 4^x + 10^x - 13.4534543^x \\<br /> f(1) > 0 \\<br /> f(2) < 0 \\[/tex]
So by the intermediate value theorem, there should exist an [tex]x_0 \in [1,2][/tex]such that
[tex] f (x_0) = 0 [/tex]
So there is a solution, I just don't think it can be solved by standard methods. These kind of things can usually only be solved in special cases or through approximations.
 
  • #11
Char. Limit said:
That is possibly a very good question...

Could we take the log of it?Maybe. Would the answer then make sense?... Maybe. The question
is... do we need an integer solution of x?

No I don't require integer solutions.

L'hopital -- Thanks, I think I am finally getting the answer I was looking for. I think starting with such a well know triple threw people off.
 

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