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Take any pythagorean triplet $(a,b,c)$, we know, by the definition that: $$a^2 + b^2 = c^2$$

But take $$a^x + b^x = c^x$$

Is $x=2$ the only possible solution $\in \Bbb R$ in this case? How can this be concluded?

I conjecture that $2$ is the only solution but I am not sure how to conclusively state this fact.

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    $\begingroup$ I have a proof but this comment box is too small to contain it. $\endgroup$ – Suzu Hirose Nov 16 '14 at 11:03
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Divide by $c^x$, and get $$\left(\frac{a}{c}\right)^x+\left(\frac{b}{c}\right)^x=1$$ The left-hand side is a decreasing function of $x$. It equals 2 when $x=0$ and approaches 0 for large $x$, so there will be only one solution.
It is greater than $2$ when $x<0$ because $a$ and $b$ are less than $c$.

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    $\begingroup$ You should also specify the case when $x\in\mathbb{R}^-$ or at least write a line about it, though it is easy to prove. Good answer +1 $\endgroup$ – Swapnil Tripathi Nov 16 '14 at 11:20
  • $\begingroup$ If I have $c^x = a^x + b^x$ and I notice $(a,b,c)$ is a pythagorean triple, can I now with full sureness say that $x$ has to be $2$? Is the reason you have stated sufficient? $\endgroup$ – Nick Nov 16 '14 at 11:31
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    $\begingroup$ It would bear stating that $(a/c),(b/c)$ are strictly between $0$ and $1$ because $(a,b,c)$ is a pythagorean triple. $\endgroup$ – hardmath Nov 16 '14 at 11:41
  • $\begingroup$ You're saying that the LHS is strictly monotonically decreasing and hence intersects $y = 1$ at only one value of $x$. Can this be verified solely on the basis of algebra or geometry? $\endgroup$ – Nick Nov 16 '14 at 23:59
  • $\begingroup$ Yes, I think that is correct. $\endgroup$ – Empy2 Nov 17 '14 at 2:16
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You should read up on Fermat's last theorem. It's not clear whether you want $x$ to be real or integer, but there is a section on exponents other than positive integers.

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  • $\begingroup$ This has little to do with FLT if the OP is also asking about real solutions $x$. $\endgroup$ – user139000 Nov 16 '14 at 11:05
  • $\begingroup$ @SuzuHirose: FTL did cross my mind but I'm a highschooler and that stuff is beyond be. I was expecting some elementary direct way of saying "yeah, x=2" $\endgroup$ – Nick Nov 16 '14 at 11:13
  • $\begingroup$ @Nick - we deleted a bunch of old comments. I've deleted three from this conversation, and David deleted at least two. $\endgroup$ – Suzu Hirose Nov 16 '14 at 11:24
  • $\begingroup$ @DavidRicherby: I understand and I have done so too myself :D $\endgroup$ – Nick Nov 16 '14 at 13:13

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