<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 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 $d<\dfrac1{42}$.
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