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$a^n + b^n = c^n$, for any integer value of n greater than two where a,b,c are positive integers.

Since this is too hard for me to solve, I tried to change the question a little. I believe Fermat put the rule "for any integer value of n greater than two" because the question would have solutions where n is any number. But even then with n being any number (except 1 and 2) I couldn't imagine of a solution. So I will ask it here if anyone knows to solve it. But if there arent any solutions that anyone knows of why did Fermat state and believe that it only has no solutions when n is an integar greater than 2.

So here is my question :

Is there solutions to $a^n + b^n = c^n$, where n is any value (positive or negative) but not 1 or 2 and a,b,c are positive integers

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    $\begingroup$ If $n$ is integer, pythagorean triplets are easy examples. If $n$ is not constrained to be an integer, then consider $(a^k)^{1/k} + (b^k)^{1/k} = ((a+b)^k)^{1/k},\quad \text{ for any }a, b, k \in \mathbb Z, k \neq 0$ $\endgroup$ – taninamdar Sep 27 '14 at 7:52
  • $\begingroup$ @taninamdar , got it. Thank you. $\endgroup$ – M.S.E Sep 27 '14 at 8:02
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If $a$, $b$, $c$ are any nonnegative numbers, you question boils down to the existence of solutions $x$ to $$u^x+v^x=1$$ where $u=a/c$ and $v=b/c$.

For $x=0$, $u^x+v^x=2>1$, so to show the existence of a solution, you only need to show there are $x$ with $u^x+v^x<1$ and apply the mean value theorem.

If $u,v<1$ you only need to pick $x$ large and positive. If $u,v>1$ you pick $x$ large and negative.

I'll leave it to you to investigate the remaining cases.

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  • $\begingroup$ Understood, Harald since now I know that solutions exists. do you know any numerical values? Or the method to solve this? $\endgroup$ – M.S.E Sep 27 '14 at 8:15
  • $\begingroup$ For some explicit examples, see the answer by @paw88789. Beyond that, though, I ca't think of any. But then I haven't spent a lot of time thinking about it either. $\endgroup$ – Harald Hanche-Olsen Sep 27 '14 at 8:28
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There are solutions for $n=-1$ and $n=-2$. For instance $10^{-1}+15^{-1}=6^{-1}$; and $15^{-2}+20^{-2}=12^{-2}$.

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  • $\begingroup$ After multiplying by the lcm of the numbers involved (squared in the second example), these reduce to the cases $n=1$ and $n=2$ (Pythagorean triples). So in a sense, this is almost cheating. But a good observation none the less. $\endgroup$ – Harald Hanche-Olsen Sep 27 '14 at 8:32
  • $\begingroup$ @Harald Hanche-Olsen: Of course it reduces...that's how I found the examples. :-) [But I don't see it as cheating--I see it as an algorithm for generating solutions in these cases.] $\endgroup$ – paw88789 Sep 27 '14 at 8:35
  • $\begingroup$ My “cheating” statement was not intended as criticism. (I tried to soften it, with “in a sense” and “almost”. Heck, I'll soften it more by upvoting your answer. There you go.) And yes, I figured that is how you found the examples. But that might not be obvious to the OP. The real challenge, of course, would be to come up with a properly non-trivial, explicit example. (My answer could also be seen as cheating in the sense of not giving an explicit answer, BTW.) $\endgroup$ – Harald Hanche-Olsen Sep 27 '14 at 9:33
  • $\begingroup$ Thanks. (I didn't really take it as criticism, which is why I sent the smiley face.) By the way, I don't think there are any 'nontrivial' examples for $n=-1$ or $n=-2$. For instance if $a^{-2}+b^{-2}=c^{-2}$, clearing denominators we obtain $(bc)^2+(ac)^2=(ab)^2$, a Pythagorean triple. $\endgroup$ – paw88789 Sep 27 '14 at 9:36
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If $n=1$, there are many solutions, right? Just pick and two positive integers for $a$ and $b$ and setting $c=a+b$ satisfies the equation. Even for $n=2$, there are infinitely many Pythagorean triples.

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  • $\begingroup$ Thank you very much, but I forgot to mention this and now edited my question. I missed to mention this while typing out the question. $\endgroup$ – M.S.E Sep 27 '14 at 8:01

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