Blistering Bananas

Firstly, note that since $\displaystyle 65536 = 4^8$ , we can say that $\displaystyle \log_465536=8$ .

Now, let us let $\displaystyle \log_2x=a$ and $\displaystyle \log_3y=b$ , such that $\displaystyle a,b \in \mathbb{R}$ .

The equation given in the problem is $\displaystyle a^4 + b^4 +8 = 8ab$ , which can be re-written as $\displaystyle \dfrac{a^4+b^4}{2} = 4(ab-1)$ .

Now, AM-GM inequality gives us $\displaystyle \dfrac{a^4+b^4}{2} \ge a^2 b^2$ . By the above equation, we can say that $\displaystyle 4(ab-1) \ge a^2b^2$ , which re-arranges to:-

$\displaystyle 0 \ge a^2 b^2-4ab+4 \implies 0 \ge (ab-2)^2$

Now, square of any real number cannot be lesser than $\displaystyle 0$ , hence, it's the equality case of the $\displaystyle \mathfrak{AM-GM}$ inequality, and therefore, $\displaystyle a=b, ab=2 \Rightarrow a=b= \sqrt2$ .

Finally, by definition of $\displaystyle \mathbb{\log}$ , $\displaystyle x=2^a=2^{\sqrt2}$ and $\displaystyle y=3^b=3^{\sqrt2}$ .

Thus, $\displaystyle xy=6^{\sqrt2} \approx 12.6$ . So, the least possible $\displaystyle N \in \mathbb{N}$ for which $\displaystyle \log_Nxy \le 1$ is $\displaystyle \boxed{13}$ $\square$