Dgdrxrt

I've heard that identities have infinitely many solutions while conditional equations have only finitely many solutions. But what about the following:

$(-1)^x=1$

This certainly isn't an identity is it? Is the problem with my definitions of conditional and identity?

Thanks!

An identity is an algebraic equality that holds true for any values of the unknowns.

An equation is an algebraic equality that holds true for some values of the unknowns. And those values are the solutions of the equation.

In your example, if $x$ is real, the solutions are all the even numbers.

The distinction between equation and identity is a bit blurred. Just avoid it and ask yourself

for what values of $x$ is $(-1)^x=1$ true?

Now realize that this is not so well defined: where is $x$ supposed to be chosen from? The integers? In this case the solution set is “the even integers”.

The reals? Oh, well, it's not possible to sensibly define $(-1)^x$ for an arbitrary real $x$. Some admit this is possible when $x=a/b$, for coprime integers $a$ and $b$, with $b$ odd. In this case the solutions are the fractions $a/b$ with $a$ even and $b$ odd.

There is another possibility: that we interpret this in the complex numbers, with the caveat that $(-1)^x$ can take on several values, possibly infinitely many. We might ask

for what complex numbers $x$ does a determination of $(-1)^x$ equal $1$?

How can we compute $(-1)^x$? It is $e^xlog(-1)$, where $log(-1)$ is any determination of the logarithm. Since $-1=e^ipi$, its logarithms have the form $(pi+2kpi)i$, for integer $k$. If $x=u+vi$, with real $u$ and $v$,
$$
x(2k+1)pi i=-(2k+1)pi v+(2k+1)pi u i
$$
and so
$$
e^x(2k+1)pi i=e^-(2k+1)pi ve^(2k+1)pi u i
$$
In order this to equal $1$, we need $v=0$. We also need
$$
(2k+1)pi u=2hpi
$$
for some integer $h$. This yields
$$
x=u+vi=u=frac2h2k+1
$$
which is the same solutions we found above.