<s>[INST] 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} $ ?
 [/INST] 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. $2^{m+3} \vert 3^{2^{m+1}}-1$.

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