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 October 22nd, 2017, 10:30 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 397 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, 05:58 PM   #2
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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, 06:34 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 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, 10:40 PM #4 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 397 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, 12:38 AM #5 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,073 Thanks: 843 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, 12:56 AM   #6
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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 03:35 AM. October 24th, 2017, 04:50 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, 09:10 PM   #8
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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 09:21 PM. October 24th, 2017, 09:22 PM #9 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 That formula is poorly written... Well, I think read... October 25th, 2017, 06:02 PM   #10
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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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post astro511 Math 2 December 26th, 2016 04:46 AM alifeee Algebra 3 November 21st, 2014 03:47 PM Arie0001 Number Theory 5 September 30th, 2013 12:46 PM cumulus.james Physics 6 February 24th, 2013 11:45 AM jaredbeach Algebra 1 November 17th, 2011 11:58 AM

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