Can we conclude that two of the variables must be $0$?

Question

Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?

I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system , is actually true.

Looking at $(a+b+c)^3$ and $(a+b+c)^2$ , using \begin{align} &(a+b+c)^3= (a+b+c)^2(a+b+c)=\\ &(1+2(ab+ac+bc))(a+b+c)= \\ &2(a+b)(a+c)(b+c)+a+b+c+2abc \end{align} and eliminating $(a+b)(a+c)(b+c)$, with $S:=a+b+c$ , I finally got $$(S-1)^2(S+2)=6abc$$

I guess this is not enough to show the above result (if it is true at all).

This question is inspired by an exercise to determine the possible values of $a+b+c$ assuming the above equations, so this question could be a duplicate, but I am not sure whether it actually is.

Answer 1

Without loss of generality assume that $a\not=0$ and $b\not=0$ such that $a^2+b^2+c^2=1$ and $a^3+b^3+c^3=1$. Then $|a|<1$, $|b|<1$ and $|c|<1$ (otherwise $a^2+b^2+c^2>1$). Therefore $$1=|a^3+b^3+c^3|\leq |a|^3+|b|^3+|c|^3<|a|^2+|b|^2+|c|^2=1$$ Contradiction.

Answer 2

Since $|a|,|b|,|c|\in[0,1]$ (because $a^2+b^2+c^2=1$), we have $$ a^2(1-a)+b^2(1-b)+c^2(1-c)=0 $$ The LHS is a sum of non-negative terms; to be zero, all the terms must be zero. This means that $$ a^2(1-a)=b^2(1-b)=c^2(1-c)=0 $$ and therefore $|a|,|b|,|c|\in\{0,1\}$. Along with $a^2+b^2+c^2=1$, we get that exactly one of them is $1$, and two must be zero: this implies the result.