[정답] $a=1,~b=-1,~c=1$

(가)에서 

$\displaystyle f(x-1)+g(x-1)=(x-1)g(x)$

$\displaystyle f(x-1)=(x-1)g(x)-g(x-1)$

이 식에 $\displaystyle f(x)=x^2+a x+b$와 $\displaystyle g(x)=x+c$을 대입하면

$\displaystyle \begin{aligned} f(x-1) &= (x-1)g(x)-g(x-1)\\&= (x-1)(x+c)-(x-1+c) \end{aligned}$

여기서 $x-1=t$로 치환하면

$\displaystyle f(t)= t^2 +ct-c$

$\displaystyle f(x)=x^2 +cx-c$

또, (나)에서 이차식 $\displaystyle f(x+2)-c g(x)$는 완전제곱식이므로

$\displaystyle \begin{aligned}f(x+2)-cg( x)&= (x+2)^2 +c(x+2)-c-c(x+c) \\&= (x+2)^2 +c -c^2 \end{aligned}$

즉, $f(x+2)-cg( x) =  (x+2)^2 +c -c^2$이 완전제곱식이므로

$\displaystyle c-c^2=0 \quad \therefore~c=0,~1$

$\displaystyle abc \neq 0$이므로 $c=1$

$\displaystyle \therefore~f(x)=x^2 +x-1$

$\displaystyle \therefore~ a=1,~b=-1, ~c=1$