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October 22nd, 2017, 11:30 PM  #1 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 303 Thanks: 25 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: 276 Thanks: 141 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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: 18,586 Thanks: 1489 
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: 303 Thanks: 25 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: 1,668 Thanks: 657 Math Focus: Wibbly wobbly timeywimey 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 
October 24th, 2017, 01:56 AM  #6  
Member Joined: Jul 2014 From: israel Posts: 76 Thanks: 3  Quote:
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:
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: 1,668 Thanks: 657 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Anyway, you said: Quote:
$\displaystyle a^{x} = \frac{1}{a^x}$, but $\displaystyle a^{1/x} = \sqrt[x]{a}$ The example I used reads: $\displaystyle a^{1/2} = \sqrt{a}$ Dan  

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