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} $ ?
 A: For the first one, induction readily gives you the answer.

For $k=1$, we have $2^3 \vert 3^{2^1} - 1$.Assume that it is true for some $k=m$ i.e. $2^{m+2} \vert 3^{2^m} - 1$ i.e. $3^{2^m}-1 = M \cdot 2^{m+2}$We have that $$3^{2^{m+1}}-1 = \left(3^{2^m}+1\right)\left(3^{2^m}-1\right) =\left(3^{2^m}+1\right) \cdot M \cdot 2^{m+2}$$But we also have that$\left(3^{2^m}+1\right) $ to be even and hence $\left(3^{2^m}+1\right) = 2 N$.Hence, we get that$$3^{2^{m+1}}-1 = 2^{m+3}MN$$i.e. $2^{m+3} \vert 3^{2^{m+1}}-1$.

A: For the second question, assume without loss of generality that $p\le q\le r$. Let $$d=1-\left(\frac1p+\frac1q+\frac1r\right)\;;$$ $d$ is rational, so let $d=\dfrac{m}n$ in lowest terms. If $$1>\frac1p+\frac1q+\frac1r>\frac{41}{42}\;,$$ then $d<\dfrac1{42}$, so $n>42$. Clearly $n\mid pqr$, so $pqr>42$, and since $4^3=64$, we must have $2\le p\le 3$.
Suppose that $p=2$; then $qr>21$, and clearly $q\ge 3$. If $q=3$, then $r>7$, so $$\frac1p+\frac1q+\frac1r\le\frac12+\frac13+\frac18=\frac{23}{24}\le\frac{41}{42}\;.$$ If $q=4$, then $r\ge 6$, and $$\frac1p+\frac1q+\frac1r\le\frac12+\frac14+\frac16=\frac{22}{24}\le\frac{41}{42}\;.$$ And if $q=5$, then $r\ge 5$, and $$\frac1p+\frac1q+\frac1r\le\frac12+\frac15+\frac15=\frac9{10}\le\frac{41}{42}\;.$$ Thus, $p\ne 2$.
If $p=3$, then $qr>14$. If $q=3$, then $r\ge 5$, and $$\frac1p+\frac1q+\frac1r\le\frac13+\frac13+\frac15=\frac{13}{15}\le\frac{41}{42}\;.$$ If $q=4$, then $r\ge 4$, and $$\frac1p+\frac1q+\frac1r\le\frac13+\frac14+\frac14=\frac5{6}\le\frac{41}{42}\;.$$ Thus, $p\ne 3$, and it is impossible to have $d<\dfrac1{42}$.
A: $$(1)$$
$$3^{2^{k}}-1=(3^2-1)(3^2+1)(3^4+1)\cdots (3^{2^{k-1}}+1)=2^3 \cdot\displaystyle\prod_{m=1}^{k-1} (3^{2^m}+1)$$
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:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\leq \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=\dfrac{41}{42}$$
