- **Number Theory**
(*http://mymathforum.com/number-theory/*)

- - **Fermat's last with inverse powers**
(*http://mymathforum.com/number-theory/342401-fermats-last-inverse-powers.html*)

Fermat's last with inverse powersDoes 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"? |

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. |

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. |

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. |

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 |

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. |

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" |

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. |

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

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 |

All times are GMT -8. The time now is 10:12 AM. |

Copyright © 2018 My Math Forum. All rights reserved.