what is the solution of m in the following equation? I have a equation  $(-1)^{2m}$=$1$ so for this $m$ has to be integer but if i write it as $((-1)^2)^m$=1 , $m$ can be rational as well. so which solution is right and why?
thanks for your help
 A: As a start consider $m = \frac12$ - then $$(-1)^{2m} = (-1)^1 = -1\ne 1$$
In general, we do not allow rational powers of negative numbers for this very reason - if $x$ is negative and $a,b \in \mathbb Q$ then in general $$x^{ab} \ne (x^a)^b$$
Some textbooks allow rational powers of negative numbers if the denominator is odd - in this case you are correct, and any $m \in \mathbb Q$ with odd denominator will be a solution.

The occurs because we define $x^{\frac12}$ to be the positive values $y$ satisfying $y^2 =x$. But $-y$ will also satisfy this equation. For example, $$\sqrt{x^2} = x$$ is only true if $x$ is non-negative.
In addition, it is not immediately clear how to define something like $(-1)^{\frac14}$. It could be any of $\pm \dfrac1{\sqrt2}\left(1\pm i\right)$ and there is no obvious way to choose which of these we should assign to be $(-1)^\frac14$.
A: Obviously, $2m$ must be an exponent that you accept as valid for a negative base. 
Irrational numbers are ruled out.
The case of integers depends on the parity. Let use denote as $e$ an even integer and $o$ an odd one: $(-1)^e=1$ and $(-1)^o=-1$.
The case of integer inverses is clear as well: $(-1)^{1/e}$ is not defined and $(-1)^{1/o}=-1$.
The case of rational needs more care as $(-1)^{p/q}$ could be interpreted both as $((-1)^p)^{1/q}$ or $((-1)^{1/q})^p$.
Anyway, 


*

*$((-1)^o)^{1/o'}=(-1)^{1/o'}=-1$ and $((-1)^{1/o'})^o=(-1)^o=-1$, so $(-1)^{o/o'}=-1$,

*$((-1)^e)^{1/o}=1^{1/o}=1$ and $((-1)^{1/o})^e=(-1)^e=1$, so $(-1)^{e/o}=1$,

*$((-1)^o)^{1/e}=(-1)^{1/e}$ and $((-1)^{1/e})^o$ are both undefined and so is $(-1)^{o/e}$.
In this sense, raising a negative to a rational power is a well defined operation.
Now it makes sense to accept $$2m=\frac eo,$$
or
$$\color{blue}{m=\frac io},$$
where $i$ is any integer.
As has been said by others, we may not use $((-1)^a)^b)=(-1)^{ab}$, so that $(-1)^{2m}=1$ and $((-1)^2)^m=1$ are two different equations, with different solutions.
