How to evaluate sum of fifth powers if sum of $n$th power is known for $n\in\{1,2,3\}$? 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:


*

*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:


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

*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.?

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}
