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 Loren October 22nd, 2017 10:30 PM

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"?

 SDK October 23rd, 2017 05:58 PM

Quote:
 Originally Posted by Loren (Post 582683) 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.

 skipjack October 23rd, 2017 06:34 PM

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.

 Loren October 23rd, 2017 10:40 PM

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.

 topsquark October 24th, 2017 12:38 AM

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

 isaac October 24th, 2017 12:56 AM

Quote:
 Originally Posted by Loren (Post 582683) 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.

 isaac October 24th, 2017 04:50 AM

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"

 Individ October 24th, 2017 09:10 PM

Quote:
 Originally Posted by skipjack (Post 582735) 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.

 Individ October 24th, 2017 09:22 PM

That formula is poorly written... Well, I think read...

 topsquark October 25th, 2017 06:02 PM

Quote:
 Originally Posted by Individ (Post 582808) $\displaystyle \frac{1}{a^x}+\frac{1}{b^x}=\frac{1}{c^x}$ 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

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