TITLE: A system of equations with 5 variables: $a+b+c+d+e=0$, $a^3+b^3+c^3+d^3+e^3=0$, $a^5+b^5+c^5+d^5+e^5=10$
QUESTION [12 upvotes]: Find the real numbers $a, b, c, d, e$ in $[-2, 2]$ that simultaneously satisfy the following relations:
$$a+b+c+d+e=0$$ $$a^3+b^3+c^3+d^3+e^3=0$$ $$a^5+b^5+c^5+d^5+e^5=10$$
I suppose that the key is related to a trigonometric substitution, but not sure what kind of substitution, or it's about a different thing.

REPLY [10 votes]: The unknowns $a,b,c,d,e$ are to be real and in the interval $[-2,2]$. This screams for the substitution $a=2\cos\phi_1$, $b=2\cos\phi_2$, $\ldots, e=2\cos\phi_5$ with some unknown angles $\phi_j,j=1,2,3,4,5$ to be made. Let's use the equations $2\cos\phi_j=e^{i\phi_j}+e^{-i\phi_j}$, $j=1,2,3,4,5$. Now
$$
0=a+b+c+d+e=\sum_{j=1}^5(e^{i\phi_j}+e^{-i\phi_j}),
$$
Using this in the second equation gives
$$
0=a^3+b^3+c^3+d^3+e^3=\sum_{j=1}^5(e^{3i\phi_j}+3e^{i\phi_j}+3e^{-i\phi_j}+e^{-3i\phi_j})
=\sum_{j=1}^5(e^{3i\phi_j}+e^{-3i\phi_j}).
$$
Using both of these in the last equation gives
$$
\begin{align}
10=a^5+b^5+c^5+d^5+e^5&=\sum_{j=1}^5(e^{5i\phi_j}+5e^{3i\phi_j}+10e^{i\phi_j}+10e^{-i\phi_j}+5e^{-3i\phi_j}+e^{-5i\phi_j})\\
&=\sum_{j=1}^5(e^{5i\phi_j}+e^{-5i\phi_j})=\sum_{j=1}^5(2\cos5\phi_j).
\end{align}
$$
This is equivalent to
$$
\sum_{j=1}^5\cos5\phi_j=5.
$$
When we know that the sum of five cosines is equal to five, certain deductions can be made :-)
This shows that there are 5 possible values for all the five unknowns, namely $2\cos(2k\pi/5)$ with $k=0,1,2,3,4$ (well, cosine is an even function, so there are only three!).  We get a solution by using each value of $k$ exactly once, because then the first two equations are satisfied (use familiar identities involving roots of unity). There may be others, but having reduced the problem to a finite search, I will exit back left.