Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?

$$(1)$$
$$3^{2^{k}}-1=(3^2-1)(3^2+1)(3^4+1)\cdots (3^{2^{k-1}}+1)=2^3 \cdot\displaystyle\prod_{m=1}^{k-1} (3^{2^m}+1)$$
Since each $3^{2^m}+1$ is even, $2$ divides each of these $k-1$ terms. 
It follows that $2^3\cdot 2^{k-1}=2^{k+2}$ divides $\displaystyle 2^3\cdot \prod_{m=1}^{k-1} (3^{2^m}+1)=3^{2^k}-1$
$$(2)$$
We wish to maximise $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r},$ without loss of generality assume $p\geq q\geq r$
Since we are dealing with three unit fractions summing to $<1$ we want $r\leq 3$. When $r=3$ the optimum solution is clearly $q=3,r=4$ which agrees with the desired inequality. 
Now let $r=2$ and we have $\dfrac{1}{p}+\dfrac{1}{q}<\dfrac{1}{2}$
Since we are now dealing with two unit fractions summing to $<1/2$ we want $q\leq 4$. When $q=4$ the optimum solution is $p=5$ which again, agrees with the inequality.
Now let $q=3$ and we have $\dfrac{1}{p}<\dfrac{1}{6}$
Here the maximum is when $p=7,$ and this last case in fact gives the greatest of all values we investigated, hence:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\leq \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=\dfrac{41}{42}$$
