Can x^n-a=0 have n real roots for n>2?

[msmith12]

In math class today, we were discussing quadratic residues, and one of the things that came up was the fact that

$$x^n-a=0$$

has n roots.

This just made me start thinking about real and complex roots. A question that I had was whether, given n>2, is it possible to have n real roots in the above eqn? If not, is there a simple proof to show why it is not possible?

 

[Muzza]

Take a = 0. One might count that as a trivial special case though...

 

[CrankFan]

http://www.cut-the-knot.org/fta/ROS2.shtml

 

[NateTG]

In math class today, we were discussing quadratic residues, and one of the things that came up was the fact that

$$x^n-a=0$$

has n roots.

This just made me start thinking about real and complex roots. A question that I had was whether, given n>2, is it possible to have n real roots in the above eqn? If not, is there a simple proof to show why it is not possible?

Well, for
$$x^n=a$$
and
$$a>0$$
The roots will be of the form:
$$\sqrt[n]{a} (\cos{\frac{k2\pi}{n} + i \sin\frac{k2\pi}{n}})$$
with $k$ ranging from $1$ to $n$.
(You can check this for yourself by, for example, multiplying, if you like).
Now, it's easy to see that this is real only if:
$$\sqrt[n]{a}=0$$
or
$$sin\frac{2k\pi}{n}=0 \Rightarrow \frac{k}{n} \in \{1,\frac{1}{2}\}$$
(There are other values, but they aren't possible for our range of k.)

From there it's easy to see that for non-zero $a$ if $n$ is even, there will be 2 real roots, and when $n$ is odd, there will be one.