An Algebra Problem from RMO

As $f(1)=0$ and $g(1)=0$ , (because $f(1)=g(1)$ )

So, 1 is a root of both equations.

Let other roots of $f(x)=0$ be $m$ and $n$ .

So, the roots of $g(x)=0$ will be $m^2$ and $n^2$ .

Therefore, $(1)(p)(q) = -c$ and $(1)(p^2)(q^2) = -a$

So, $-a=c^2$

And, $-a =p+q+1$

Squaring both sides,

$a^2=p^2 + q^2 + 1 +2(pq+p+q) = -b +2b = b$

Therefore, $b=c^4$ .

And as $f(1)=0$ , we have,

$1+c-c^2+c^4 = 0$ $(c+1)(c^3-c^2+1=0)$

And, as $a,b,c$ are integers, $\boxed{c=-1}$

So, $\boxed{b=1}$ and $\boxed{a=-1}$

Therefore, $a^{2013}+b^{2013}+c^{2013}= \boxed{-1}$

So, $|a^{2013}+b^{2013}+c^{2013}|= \boxed{1}$

Use newtons sum, let $\alpha,\beta, \gamma$ be roots of $p(x)$ let $\alpha^2 ,\beta^2, \gamma^2$ be roots of $g(x)$ we know $\alpha^2 +\beta^2+ \gamma^2=a^2 -2b$ $\alpha^2 +\beta^2+ \gamma^2=-b$ we get $b=a^2$ again $\alpha \beta \gamma = -c$ $\alpha^2 \beta^2 \gamma^2 =-a$ so $a=-c^2$ since f(1)=0 $a+b+c=-1$ $-c^2 +c^4+c-1=0$ and the only integral solution is $+1$ so, $a= -1, c=+1,b=+1$ solve and get $\boxed{1}$