How to find all eigenvalues of this monster (complex) matrix? Let 
$$a=\cos(\frac25\pi)+i\sin(\frac25\pi)$$
and 
$$A=\begin{pmatrix}1&1&1&1&1\\1&a&a^2&a^3&a^4\\1&a^2&a^4&a^6&a^8\\1&a^3&a^6&a^9&a^{12}\\1&a^4&a^8&a^{12}&a^{16} \end{pmatrix}$$
Then what I need to do is find all eigenvalues of $A$.
To be honest I don't know where to start even though I can sense that there is definitely something special about $A$. So I'm sure I'm not supposed to use brutal force calculating the characteristic polynomial. But there is just nothing else I can think up. Can you help me? Giving little bit of hint (which I hope will not be too obscure to understand for a college freshman) will be specially appreciated. Thanks a lot! 
 A: $a$ is a root of unity. You can see that if you  write $a=e^{\frac {2}{5} \pi i}$. So $r_0=a^0$, $_1=a^1$ ... and $r_4=a^4$ are the 5-roots of unity. The 5-roots of unity are in the form $e^{\frac {2}{5} \pi ik}$ for $k=0,1,2,3,4$. So $a^6=a,a^8=a^3,a^9=a^4,a^{12}=a^2,a^{16}=a$. Every line has the elements $1,a,a^2,a^3,a^4$ in different positions. This is just a hint.
A: Hint:
I have not had time to finish it, but here is a way to find all the possible candidate eigenvalues.
If we label the rows and columns by $i,j=0,1,2,3,4$ then $A_{ij} = a^{ij}$ and 
$$A^2_{ij} = \sum_{k=0}^4 A_{ik}A_{kj} = 1 + a^{(i+j)} + a^{2(i+j)} + a^{3(i+j)} + a^{4(i+j)} = \left\{\begin{array}{cc}5 & \text{if}~~~ 5|(i+j)\\0 & \text{otherwise}\end{array}\right.$$
where we have used $a^5 = 1$ and $\sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1}$. From this it follows that
$$A^2 = \left(\matrix{5 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 5\\0 & 0 & 0 & 5 & 0\\0 & 0 & 5 & 0 & 0\\0 & 5 & 0 & 0 & 0}\right)$$
It also follows that $A^4 = 25 I$. If $\lambda$ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^2$ (though the converse is not always true). The eigenvalues of $A^2$ are $\pm 5$ so the only possible eigenvalues of $A$ are
$$\{\sqrt{5},-\sqrt{5},i\sqrt{5},-i\sqrt{5}\}$$
It remains to show that these are indeed all eigenvalues of $A$.
