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While studying Fermat's Last Theorem and Pythagorean triples, the following question occurred to me: For the equation $a^n+b^n=c^n$, where $n$ is a negative integer, a) does a solution exist, and b) if solutions exist, is there some analog to Fermat's Last Theorem for these parameters? I have made a few passing attempts at finding a solution and have come up empty handed, though I am no great mathematical mind and would not be shocked if I have missed even obvious answers. Thank you in advance for your aid.

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    $\begingroup$ You can multiply by $(abc)^n$ to get back into the case of positive powers. $\endgroup$ – Adam Hughes Jul 11 '14 at 5:02
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    $\begingroup$ $3^{-1}+6^{-1} = 2^{-1}$ $\endgroup$ – JimmyK4542 Jul 11 '14 at 5:02
  • $\begingroup$ @JimmyK4542 thank you for the solution. Now , does one exist for, say, $n<-2$? $\endgroup$ – Gotthold Jul 11 '14 at 5:11
  • $\begingroup$ no, you'd have $$bc+ac=ab$$ $\endgroup$ – Adam Hughes Jul 11 '14 at 5:12
  • $\begingroup$ @AdamHughes I'm sorry sir. $\endgroup$ – Gotthold Jul 11 '14 at 5:15
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Let's suppose that $a^{-n}+b^{-n} = c^{-n}$ for some positive integers $a,b,c,n$.

Then, multiply both sides by $a^nb^nc^n$ to get $b^nc^n + a^nc^n = a^nb^n$, i.e. $(bc)^n+(ac)^n=(ab)^n$.

If $n \ge 3$, then this contradicts Fermat's Last Theorem. Hence, there are no solutions for $n \ge 3$.

For $n = 1$, we have several solutions, one of which is $3^{-1}+6^{-1} = 2^{-1}$.

For $n = 2$, we have several solutions, one of which is $15^{-2}+20^{-2} = 12^{-2}$.

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Fermat's theorem actually asks the above question for positive integers greater than two; so are you asking for negative $n$ less than negative two?

Anyways, to answer your question consider the problem for $n = -1$. We have $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}. $$ This means that if we have $(a+b) \vert ab$ then we can simply define $c=\frac{ab}{a+b}$. There exist many formulas for finding two distinct integers such that their sum divides their product (see link: Necessary and sufficient conditions for the sum of two numbers to divide their product).

Ex: let $a=10$, $b=15$. We have $$ \frac{1}{10}+\frac{1}{15} = \frac{1}{6}. $$

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  • $\begingroup$ Sir, thank you. Yes, I was asking about n values less than -2. $\endgroup$ – Gotthold Jul 11 '14 at 5:18
  • $\begingroup$ For $n<-2$ you get $$(bc)^n+(ac)^n=(ab)^n$$ which is not solvable by Fermat's last theorem. $\endgroup$ – Adam Hughes Jul 11 '14 at 5:23

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