TITLE: Largest $k$ for which $a^k +b^k \leq c$
QUESTION [7 upvotes]: I have an inequality of the form $$a^k +b^k \leq c$$ with $a,b,c,k \in\mathbb{Z^+}$. For known $a,b,c$ I want to find out the largest $k$ for which this inequality holds. I am able to write a program that does this for me, but cannot come up with any clever way to do it analytically. Is there a good way to do this?

REPLY [9 votes]: Without loss of generality, let $b > a$. Notice that 
$$b^k \leq a^k + b^k \leq 2 b^k$$
So if you want that $a^k + b^k \leq c < a^{k+1} + b^{k+1}$, then it follows that
$$b^k \leq c < 2 b^{k+1}$$
Or, equivalently, 
$$k \leq \log_b(c) < k + 1 + \log_b(2)$$
Finding the right value of $k$ should be easy now. If $a = b = 1$ the problem is trivial, while for $b \geq 2$ you have $\log_b(2) \leq 1$ such that
$$k \leq \log_b(c) < k + 2$$
As the only integral values of $k$ satisfying this relation are $k = \lfloor \log_b(c) \rfloor$ and $k = \lfloor \log_b(c) \rfloor - 1$, you only need to check these two values as possible solutions.