12
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ +1: Where does your problem come from? What about the other power sums, $p_2$ and $p_4$? $\endgroup$
    – draks ...
    Commented Jun 25, 2012 at 13:57
  • $\begingroup$ Is there any reason to believe there will be a unique solution? With three equations and five variables, one would naively expect many solutions. $\endgroup$ Commented Jun 25, 2012 at 15:07

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ So NullUser was on the right track. +1 for that. Another +1 to Mark Bennet, for my first idea was also to use Newton(-Girard) identities. $\endgroup$ Commented Jun 25, 2012 at 16:14
  • $\begingroup$ then i have (+1) for you. :-). Your exposure is very clear.Thanks! $\endgroup$ Commented Jun 25, 2012 at 16:16
7
$\begingroup$

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 $a,b,c,d,e$.

$\endgroup$
4
  • 1
    $\begingroup$ Nice. But those aren't all real numbers in $[-2,2]$. $\endgroup$
    – Zander
    Commented Jun 25, 2012 at 15:41
  • $\begingroup$ Shoot =\. I read that as $|z|< 2$ and forgot about the whole real numbers requirement. I'll leave the solution up and make a comment about that. $\endgroup$
    – nullUser
    Commented Jun 25, 2012 at 15:43
  • 2
    $\begingroup$ Yes, this does work taking the real parts, but they need to be scaled by 2 instead of $2^{1/5}$. $\endgroup$
    – Zander
    Commented Jun 25, 2012 at 15:54
  • $\begingroup$ @Zander Can you explain why we can just take real parts? It is not true that $[Re(e^{i\theta})]^3 = Re(e^{i3 \theta})$. $\endgroup$
    – Calvin Lin
    Commented Sep 17, 2013 at 22:15
3
$\begingroup$

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]

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .