A system of equations with 5 variables: $a+b+c+d+e=0$, $a^3+b^3+c^3+d^3+e^3=0$, $a^5+b^5+c^5+d^5+e^5=10$ 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.
 A: EDIT: as was pointed out, this solution is invalid since the answers are complex numbers.
EDIT 2: Actually this does work, take the real parts of all the complex solutions
I can't really explain why I thought to do this, but it worked, probably because you mentioned it should have a trig solution.
We know that the sum of the 5th roots of unity is 0, i.e. that
$$
\sum_{k=0}^4e^{i2\pi k/5} =0.
$$
What happens if we consider the powers? Turns out that
$$
\sum_{k=0}^4(e^{i2\pi k/5})^3 = 0
$$
too (to see this, note that $x\longmapsto x^3$ is an automorphism since the order of the group is 5), and
$$
\sum_{k=0}^4(e^{i2\pi k/5})^5 = \sum_{k=0}^41 = 5.
$$
with this knowledge it is simple; scale all the variables by $2^{1/5}$.
Take $\{ 2^{1/5}e^{i2\pi k/5} \}_{k=0}^4$ for $a,b,c,d,e$.
A: This is a long suggestion of a way to get started rather than a proper answer:
If you put $a,b,c,d,e \text { as roots of a quintic polynomial } x^5-p_1x^4+p_2x^3-p_3x^2+p_4x-p_5=0$
$ \text {Noting that } p_1=0$ substitute in the five values and add the equations, using $s_r$ to denote the sum of the $r$th powers of the roots you obtain:
$$s_5-p_1s_4+p_2s_3-p_3s_2+p_4s_1-5p_5=0$$
Subsituting known values this becomes:
$$10-p_3s_2-5p_5=0$$
$0 \leq s_2 = a^2+b^2+c^2+d^2+e^2 \leq 20$
$p_3$ is the sum of products like $abc$ - distinct roots taken three at a time.
$p_5 = abcde$
The constraint suggests using inequalities from this point to bound the possibilities. [The fact that the question has been asked is sometimes suggestive too, depending on who asked it]
A: 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.
