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Suppose we have numbers $a,b,c$ which satisfy the equations $$a+b+c=3,$$ $$a^2+b^2+c^2=5,$$ $$a^3+b^3+c^3=7.$$

How can I find $a^5 + b^5 + c^5$?

I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:

$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$

I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.

Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$: $$\overline{f}^G = \frac{29}{3}$$

How can this remainder be of use to me?

Thanks. (Note: I am using Macaualay2)

Patrick Shambayati
  • 1,523

3 Answers3

4

You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities

In general if you have $n$ variables $x_1\ldots.x_n$, define the polynomials $$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$ and \begin{align} e_0(x_1, \ldots, x_n) &= 1,\\ e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\ e_2(x_1, \ldots, x_n) &= \textstyle\sum_{1\leq i<j\leq n}x_ix_j,\\ e_n(x_1, \ldots, x_n) &= x_1 x_2 \cdots x_n,\\ e_k(x_1, \ldots, x_n) &= 0, \quad\text{for}\ k>n.\\ \end{align} Then \begin{align} p_1 &= e_1,\\ p_2 &= e_1p_1-2e_2,\\ p_3 &= e_1p_2 - e_2p_1 + 3e_3 ,\\ p_4 &= e_1p_3 - e_2p_2 + e_3p_1 - 4e_4, \\ & {}\ \ \vdots\\ \\ e_0 &= 1,\\ e_1 &= p_1,\\ e_2 &= \frac{1}{2}(e_1 p_1 - p_2),\\ e_3 &= \frac{1}{3}(e_2 p_1 - e_1 p_2 + p_3),\\ e_4 &= \frac{1}{4}(e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4),\\ & {} \ \ \vdots \end{align}

In your case you have only 3 variables. Using the formulas above compute \begin{align} e_1 &=p_1=3, \\ e_2 &=2,\\ e_3 &=-\frac{2}{3}\\ e_4 &=0\\ e_5 &=0\\ \end{align} Then compute $p_4=9$ and $$p_5=\frac{29}{3}.$$

Sfarla
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0

Using just Macaulay2, you can do the following

Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
               ReesAlgebra, TangentCone

i1 : R=QQ[a,b,c]

o1 = R

o1 : PolynomialRing

i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7)

                            2    2    2       3    3    3
o2 = ideal (a + b + c - 3, a  + b  + c  - 5, a  + b  + c  - 7)

o2 : Ideal of R

i3 : S=R/i1

o3 = S

o3 : QuotientRing

i4 : phi=map(S,R)

o4 = map(S,R,{- b - c + 3, b, c})

o4 : RingMap S <--- R

i6 : use R

o6 = R

o6 : PolynomialRing

i7 : phi(a^5+b^5+c^5)

     29
o7 = --
      3

o7 : S

(I deleted i5 and o5 as I made a typo in the input there)

Jürgen Böhm
  • 3,105
0

Given $a+b+c=3$ and $a^2+b^2+c^2 =5$ and $a^3+b^3+c^3=7$

Using $$ab+bc+ca = \frac{1}{2}\left[(a+b+c)^2-(a^2+b^2+c^2)\right] = 2$$

and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]=9$$

So $$7-3abc=9\Rightarrow abc=-\frac{2}{3}$$

Now Let $(t-a)\;,(t-b)\;,(t-c)$ be the root of cubic equation in terms of $t\;,$ Then

$$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$

So $$t^3-3t^2+2t+\frac{2}{3}=0\Rightarrow t^4-3t^3+2t^2+\frac{2}{3}t=0......(1)$$

So $$\sum a^4-3\sum a^3+2\sum a^2+\frac{2}{3}\sum a=0$$, Where $\sum a^{n} = a^n+b^n+c^n\;,$ for $n=1,2,3,4,5$

So $$\sum a^4-3(7)+2\cdot 5+\frac{2}{3}\cdot 3=0\Rightarrow \sum a^4=9$$

Now $$t^5-3t^4+2t^3+\frac{2}{3}t^2=0$$ from equation $(1)$

So $$\sum a^5-3\sum a^4+2\sum a^3+\frac{2}{3}\sum a^2=0$$

So $$\sum a^5-3(9)+2(7)+\frac{2}{3}\cdot 5=0\Rightarrow \sum a^5 = \frac{29}{3}$$

juantheron
  • 53,015