Function No. 2013

Note that $g(1) = f(1) = 0$ , so 1 is the root of both $g(x) and f(x)$ .
Let p and q are other two roots of f(x), so $p^2$ and $q^2$ are other two roots of $g(x)$ . We then get $-c = pq$ and $-a = p^2 q^2$ , so $a = -c^2$ .
Also $(-a)^2 = ( p + q+ 1)^2 = p^2 + q^2 + 1 + 2( pq + q + p ) = b$ .
Therefore $b = c^4$ . Since $f(1) = 0, c +1 - c^2 + c^4 = 0$ . Factorizing, we get $( c + 1) ( c^3 - c^2 + 1 ) = 0$ .
Note that $c^3 - c^2 + 1 = 0$ has no integer root hence, $c = -1, b = 0 , a = -1$ . So, $a^{2013} + b^{2013 }+ c^{2013} = -1$ .

We know that both $f(1)=0$ and $g(1)=0$ because $1^2=1$ . Substituting $1$ into either polynomial gives the same result, relating $a$ , $b$ , and $c$ : $a+b+c+1=0$ . Then I use synthetic division to partially factor each polynomial as $f(x)=(x-1)(x^2+(a+1)x+a+b+1)$ and $g(x)=(x-1)(x^2+(b+1)x+b+c+1)$ . Using the relationship $a+b+c+1=0$ , I can write $g(x)=(x-1)(x^2+(b+1)x-a)$ , eliminating the variable $c$ . Let $\alpha$ and $\beta$ be the other roots of $f(x)$ --other than the known root $x=1$ . Then through applying Vieta's formulas to both $f(x)$ and $g(x)$ , I get the four equations: $\alpha+\beta=-a-1,\alpha\beta=a+b+1,\alpha^2+\beta^2=-b-1,\alpha^2\beta^2=-a$ . I first square the second equation, so that I can equate it with the fourth: $\alpha^2\beta^2=-a=a^2+b^2+2ab+2a+2b+1$ .Then I square the first, multiply the second by $2$ , subtract the resulting equations, and equate it with the third equation to get: $\alpha^2+\beta^2=a^2-2b-1=-b-1$ , which implies $a^2=b$ . Then after substituting $a^2=b$ into $a^2+b^2+2ab+3a+2b+1=0$ , and simplifying, I get $a^4+2a^3+3a^2+3a+1=0$ . There is only one integer root of this polynomial (cf. https://brilliant.org/wiki/rational-root-theorem/), which leads to an answer of $a=-1, b=1, c=-1$ . Hence, $a^{2013}+b^{2013}+c^{2013}=(-1)^{2013}+(1)^{2013}+(-1)^{2013}=-1$ .