Mock AIME II 2012 Problems/Problem 11


There exist real values of $a$ and $b$ such that $a+b=n$, $a^2+b^2=2n$, and $a^3+b^3=3n$ for some value of $n$. Let $S$ be the sum of all possible values of $a^4+b^4$. Find $S$.


First, if $n=0$, then $a^2+b^2=0\implies a=b=0$. We now assume that $n\ne 0$. Now, note that $2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab$. Also, we have $(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies$ $2n+2(2n-3)=n^2\implies n^2-6n+6=0$.

Next, $3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)$ $\implies a^4+b^4=-n^2+6n$. But we know $n^2-6n+6=0$, so $-n^2+6n=6$.

Since the only possible values of $a^4+b^4$ are $0$ and $6$, our final answer is $\boxed{006}$.

(It is easy to check that there exists $a, b, n$ satisfying the equations.)