Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the real numbers $a, b, c, d, e$ in $[-2, 2]$ that simultaneously satify 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.

share|improve this question
    
+1: Where does your problem come from? What about the other power sums, $p_2$ and $p_4$? –  draks ... Jun 25 '12 at 13:57
    
Is there any reason to believe there will be a unique solution? With three equations and five variables, one would naively expect many solutions. –  David Speyer Jun 25 '12 at 15:07
add comment

3 Answers 3

up vote 8 down vote accepted

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.

share|improve this answer
    
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. –  Jyrki Lahtonen Jun 25 '12 at 16:14
    
then i have (+1) for you. :-). Your exposure is very clear.Thanks! –  Chris's sis Jun 25 '12 at 16:16
add comment

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

share|improve this answer
1  
Nice. But those aren't all real numbers in $[-2,2]$. –  Zander Jun 25 '12 at 15:41
    
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. –  nullUser Jun 25 '12 at 15:43
2  
Yes, this does work taking the real parts, but they need to be scaled by 2 instead of $2^{1/5}$. –  Zander Jun 25 '12 at 15:54
    
@Zander Can you explain why we can just take real parts? It is not true that $[Re(e^{i\theta})]^3 = Re(e^{i3 \theta})$. –  Calvin Lin Sep 17 '13 at 22:15
add comment

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]

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.