The positive integers a, b, and c satisfy a^(6) + b^(2) + c^(2) = 2011. Find a + b + c.

Question

anonymous · Answer

It's a question from a previous math competition.  I've been trying to find a rigorous solution for it...

rational · Answer

anonymous · Answer

I've actually already seen that link.  I don't understand where he pulled some of those numbers. It's quite vague

rational · Answer

First observation :  $a \le 3$ as   $4^6 \gt   2011$

anonymous · Answer

Your correct, and that is about as far I have gotten. figuring out the other two have proved to be very difficult.

rational · Answer

rational · Answer

that link is all we need i think...

rational · Answer

3 cases :$ a = 1, 2, 3$

for each case, apply the theorem in that link to check if $ 2011-a^2$ can be expressed as sum of two squares.

rational · Answer

$2011-a^6$  *

anonymous · Answer

a=3 b=21 and c= 29
hence a+b+c=53