HMMT February

Let $\sigma_{1}=a+b+c, \sigma_{2}=ab+bc+ca$ and $\sigma_{3}=abc$ be the three elementary symmetric polynomials. Since $a^{3}+b^{3}+c^{3}$ is a symmetric polynomial, it can be written as a polynomial in $\sigma_{1}, \sigma_{2}$ and $\sigma_{3}$. Now, observe that $\sigma_{1}=0$, and so we only need to worry about the terms not containing $\sigma_{1}$. By considering the degrees of the terms, we see that the only possibility is $\sigma_{3}$. That is, $a^{3}+b^{3}+c^{3}=k\sigma_{3}$ for some constant $k$. By setting $a=b=1, c=-2$, we see that $k=3$. By similar reasoning, we find that $a^{5}+b^{5}+c^{5}=h\sigma_{2}\sigma_{3}$ for some constant $h$. By setting $a=b=1$ and $c=-2$, we get $h=-5$. So, we now know that $a+b+c=0$ implies $$a^{3}+b^{3}+c^{3}=3abc \quad \text { and } \quad a^{5}+b^{5}+c^{5}=-5abc(ab+bc+ca)$$ Then $a^{3}+b^{3}+c^{3}=a^{5}+b^{5}+c^{5}$ implies that $3abc=-5abc(ab+bc+ca)$. Given that $a, b, c$ are nonzero, we get $ab+bc+ca=-\frac{3}{5}$. Then, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)=\frac{6}{5}$.