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

[Diffy]

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.

 

[Char. Limit]

FLT states that the only way all of those solutions could be integers is if x=2.

 

[Diffy]

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.

 

[D H]

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?

 

[Diffy]

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?

 

[l'Hôpital]

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 a^x + b^x = c^x 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.

 

[CaffeineJunky]

Fermat-Wiles theorem. Give the guy credit who proved it along with the one who conjectured it.

 

[Diffy]

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...

 

[Char. Limit]

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?

 

[l'Hôpital]

<br /> \begin{array}{1}<br /> f(x) = 4^x + 10^x - 13.4534543^x \\<br /> f(1) &gt; 0 \\<br /> f(2) &lt; 0 \\<br />
So by the intermediate value theorem, there should exist an x_0 \in [1,2]such that
<br /> f (x_0) = 0 <br />
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.

 

[Diffy]

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.