Berkeley Math Circle Monthly Contest 3

Solution:
We have that $b=\left(a^{2}+1\right)^{3}-1$, $c=\left(b^{2}+1\right)^{3}-1$, and $a=\left(c^{2}+1\right)^{3}-1$. By direct substitution we derive that $a$ satisfies the following polynomial equation of degree 216:
$$\left(\left(\left(\left(\left(a^{2}+1\right)^{3}-1\right)^{2}+1\right)^{3}-1\right)^{2}+1\right)^{3}-(a+1)=0$$
We observe that the polynomial can be rewritten as
$$a^{216}+c_{215} a^{215}+\cdots+c_{2} a^{2}-a=0$$
for some integers $c_{2}, \ldots, c_{215}$. Hence by the Rational Root Theorem, if $a \neq 0$ then it follows that $a= \pm 1$. So $a \in\{-1,0,1\}$. Similarly, $b, c \in\{-1,0,1\}$ as well.

But if $a= \pm 1$, then we have $b=(1+1)^{3}-1=7$, which is impossible. Hence only $a=0$ can occur. Thus $a=b=c=0$.