Can any of these polynomials be a square? It's in $\mathbb C[x]$ and they have the form
$$1+x+x^2+\dots +x^n$$
Obviously $n$ can't be odd. I can prove it for any specific polynomial via GCD with the derivative, but how to prove it for all $n$ at once?
 A: Hint : 
$$1 + x+ x^2 +\cdots +x^n=\frac{x^{n+1}-1}{x-1}$$
A: If it were a square then it'd have at least one repeated root $\;\alpha\;$ , and this can happen iff also its derivarive vanishes at $\;\alpha\;$, but
$$f(x)=1+x+\ldots+ x^n\implies f'(x)=1+2x+3x^2+\ldots+nx^{n-1} $$
so we have
$$\begin{align*}&I\;\;\;\;1+\alpha+\alpha^2+\ldots+\alpha^n=0\\
&II\;\;1+2\alpha+3\alpha^2+\ldots+n\alpha^{n-1}=0\end{align*}$$
Substracting 1 from 2 in the above:
$$\alpha+2\alpha^2+3\alpha^3+\ldots+(n-1)\alpha^{n-1}-\alpha^n=0$$
and since clearly $\;\alpha\neq 0\;$ :
$$1+2\alpha+3\alpha^2+\ldots+(n-1)\alpha^{n-2}-\alpha^{n-1}=0$$
equating this last equation with $\;II\;$ above we get:
$$\alpha^{n-1}=0\;,\;\;\text{contradiction}$$
A: To elaborate on Digital Brains Hint,
Consider the specific example of,
$$ 1 + x + x^2 = \frac{x^3-1}{x-1} \Rightarrow (x-1)(1+x+x^2) = x^3-1$$
The roots of the last expression on  the right are the cubed roots of $1$ which are $1,\frac{-1-i\sqrt{3}}{2}, \text{ and } \frac{-1+i\sqrt{3}}{2}$; all of these roots have a multiplicity of one. The root $x=1$ clearly corresponds to $(x-1)$ on the left so we throw that away. The other two complex roots are the roots of the original polynomial $1+x+x^2$, since they only occur with degree one we conclude that this polynomial is not a square.
Generalizing this is pretty easy. The roots of any of these cyclotomic polynomials are just the roots of unity that aren't equal to one. So we can extend the proof to show that all such polynomials are not squares.
A: Following the hint:
$$1+x+x^2+\dots+x^n=\frac{x^{n+1}-1}{x-1}$$
We relate the zeros of the first polynomial to the second numerator. Now if you recognize the second polynomial as the one giving you the $n+1$ distinct roots of unity you are finished. On the other hand if you derive $x^{n+1}-1$ you get $(n+1)x^n$ and you can see that the polynomial $x^{n+1}-1$ doesn't have multiple zeros.
