My Math Forum Fermat's last with inverse powers

 Number Theory Number Theory Math Forum

 October 22nd, 2017, 11:30 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 390 Thanks: 27 Math Focus: Number theory Fermat's last with inverse powers Does a^(x^-1) + b^(x^-1) = c^(x^-1) hold true for any integers {a, b, c, x}>1 where "a" does not equal "b"?
October 23rd, 2017, 06:58 PM   #2
Senior Member

Joined: Sep 2016
From: USA

Posts: 502
Thanks: 280

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by Loren Does a^(x^-1) + b^(x^-1) = c^(x^-1) hold true for any integers {a, b, c, x}>1 where "a" does not equal "b"?
I think this says $a^{-x} + b^{-x} = c^{-x}$?

In this case, you are interested when $x \in \mathbb{N}$ which you can easily see has solutions whenever there are rational solutions of
$s^x + t^x = 1$

It is fairly easy to see this has solutions for any $x \in \mathbb{N}$ which implies the "inverse fermat's" is both false, and a lot more boring.

 October 23rd, 2017, 07:34 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,866 Thanks: 1833 As posted, it would be $\displaystyle a^{x^{-1}} + b^{x^{-1}} = c^{x^{-1}}$, i.e. $\displaystyle a^{1/x} + b^{1/x} = c^{1/x}$, which clearly has integer solutions.
 October 23rd, 2017, 11:40 PM #4 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 390 Thanks: 27 Math Focus: Number theory Is there a simple rule for generating solutions to this "inverse Fermat" equation? Please tell me what I am missing that gives many integer {a, b, c, x} for unequal a and b; e.g. some examples of four integers that obey the equation.
 October 24th, 2017, 01:38 AM #5 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,908 Thanks: 773 Math Focus: Wibbly wobbly timey-wimey stuff. I'll do an example of x = 2 and you can generalize yourself. Let x = 2. Then $\displaystyle a^{1/2} + b^{1/2} = c^{1/2}$ Let $\displaystyle a = p^2$, $\displaystyle b = q^2$, and $\displaystyle c = r^2$, so $\displaystyle a^{1/2} + b^{1/2} = c^{1/2} \implies p + q = r$ So, specifically, let p = 2, q = 3 giving r = 5. Then a = 4, b = 9 and c = 25. You can generate any number of solutions this way, for any x desired. -Dan Thanks from Loren and Omne Bonum
October 24th, 2017, 01:56 AM   #6
Member

Joined: Jul 2014
From: israel

Posts: 76
Thanks: 3

Quote:
 Originally Posted by Loren Does a^(x^-1) + b^(x^-1) = c^(x^-1) hold true for any integers {a, b, c, x}>1 where "a" does not equal "b"?

a^(x^-1) + b^(x^-1) = c^(x^-1)

a = A^x
b = B^x
c = C^x

a^(x^-1) + b^(x^-1) = c^(x^-1)
(A^x)^(1/x) + (B^x)^(1/x) + (C^x)^(1/x)
A + B = C

I guess there are more solutions than this, but I think this is the most trivial one.

Last edited by skipjack; October 24th, 2017 at 04:35 AM.

 October 24th, 2017, 05:50 AM #7 Member   Joined: Jul 2014 From: israel Posts: 76 Thanks: 3 i also just wanted to state the obvious if it wasn't clear by now this will work for any given integer x>1 where "a" does/doesnt equal "b"
October 24th, 2017, 10:10 PM   #8
Member

Joined: Jul 2015
From: tbilisi

Posts: 30
Thanks: 8

Quote:
 Originally Posted by skipjack As posted, it would be $\displaystyle a^{x^{-1}} + b^{x^{-1}} = c^{x^{-1}}$, i.e. $\displaystyle a^{1/x} + b^{1/x} = c^{1/x}$, which clearly has integer solutions.
$\displaystyle \frac{1}{a^x}+\frac{1}{b^x}=\frac{1}{c^x}\$

Then.

$\displaystyle \frac{{(bc)}^x}{{(abc)}^x}+\frac{{(ac)}^x}{{(abc)} ^x}=\frac{{(ab)}^x}{{(abc)}^x$

Then.

$\displaystyle {(bc)}^x+{(ac)}^x={(ab)}^x$

Then.

$\displaystyle k^x+t^x=q^x$

For the case when $\displaystyle x>2$ No solutions. The same, Fermat's equation written differently.

Last edited by Individ; October 24th, 2017 at 10:21 PM.

 October 24th, 2017, 10:22 PM #9 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 That formula is poorly written... Well, I think read...
October 25th, 2017, 07:02 PM   #10
Math Team

Joined: May 2013
From: The Astral plane

Posts: 1,908
Thanks: 773

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Individ $\displaystyle \frac{1}{a^x}+\frac{1}{b^x}=\frac{1}{c^x}$ Then. $\displaystyle \frac{(bc)^x}{(abc)^x}+\frac{(ac)^x}{(abc)^x}= \frac{(ab)^x}{(abc)^x}$ Then. $\displaystyle {(bc)}^x+{(ac)}^x={(ab)}^x$ Then. $\displaystyle k^x+t^x=q^x$ For the case when $\displaystyle x>2$ No solutions. The same, Fermat's equation written differently.
I was able to fix the code.

Anyway, you said:
Quote:
 $\displaystyle \frac{1}{a^x}+\frac{1}{b^x}=\frac{1}{c^x}$
This is not the same as $\displaystyle a^{1/x} + b^{1/x} = c^{1/x}$.

$\displaystyle a^{-x} = \frac{1}{a^x}$, but $\displaystyle a^{1/x} = \sqrt[x]{a}$

The example I used reads: $\displaystyle a^{1/2} = \sqrt{a}$

-Dan

 Tags fermat, inverse, powers

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post astro511 Math 2 December 26th, 2016 05:46 AM alifeee Algebra 3 November 21st, 2014 04:47 PM Arie0001 Number Theory 5 September 30th, 2013 01:46 PM cumulus.james Physics 6 February 24th, 2013 12:45 PM jaredbeach Algebra 1 November 17th, 2011 12:58 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top