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In a book of mine, I found two problems I couldn't solve. Namely:

Let $a,b,c$ satisfies $a+b+c=3,a^2+b^2+c^2=5, a^3+b^3+c^3=7$. There are three questions:

  1. Find $a^4+b^4+c^4$. There was a solution that the value of the sum of fourth powers is $9$.

But I was unable to solve the following two questions:

  1. Does the pattern continue? Does $a^5+b^5+c^5=11$?

  2. Show that there are no solutions for $a,b$, and $c$ in the real numbers, but there are solutions in the complex numbers.

I tried to solve 2. by Sage:

sage: factor((a+b+c)^5-a^5-b^5-c^5)
5*(a^2 + a*b + b^2 + a*c + b*c + c^2)*(a + b)*(a + c)*(b + c)

I have no idea how to do the part 3. Can anyone show me how to solve 2. and 3.?

studying
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  • Use Newton's identities to calculate the symmetric polynomials in variables $a,b,c$, then use those to calculate the sums of higher powers. See an example worked out step by step in this answer to a related question. For part 3, once you have the values of the symmetric polynomials, you know the cubic equation which has $a,b,c$ as roots, and can determine whether the roots are all real or not. – dxiv Oct 08 '16 at 18:33
  • The idea behind divx's comment is that you can build $a^4+b^4+c^4$ by starting with $(a+b+c)^4$ and subtracting correct amounts of $(a+b+c)(a^3+b^3+c^3)$, $(a+b+c)^2(a^2+b^2+c^2)$ and other such products. – Arthur Oct 08 '16 at 19:02

1 Answers1

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We have \begin{align} a^2+b^2+c^2&=(a+b+c)^2-2ab-2bc-2ca\\ 5&=9-2(ab+bc+ca)\\ ab+bc+ca&=2\\ (ab+bc+ca)^2&=4\\ a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)&=4\\ a^2b^2+b^2c^2+c^2a^2+6abc&=4\qquad\text{since }a+b+c=3\\ \frac{(a^2+b^2+c^2)^2-(a^4+b^4+c^4)}{2}+6abc&=4\tag{1} \end{align} On the other hand, from the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ we have \begin{align} 7-3abc&=(3)\left(5-2\right)\\ abc&=-\frac{2}3 \end{align} plugging it into $(1)$ we get \begin{align} \frac{5^2-(a^4+b^4+c^4)}{2}+6\left(-\frac{2}3\right)&=4\\ 25-(a^4+b^4+c^4)-8&=8\\ a^4+b^4+c^4&=9 \end{align}

Ángel Mario Gallegos
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