Let P be the set of all prime numbers and let S = {t :2^t -1 is prime}. Prove that S is a subset of P.

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$$\begin{gathered} \mathbf{Prime}\mathbf{ }\mathbf{Number}\mathbf{=}\mathbf{Prime}\mathbf{ }\mathbf{Number}\mathbf{\times }\mathbf{1}\mathbf{ }\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{ }\mathbf{it}\mathbf{ }\mathbf{has}\mathbf{ }\mathbf{only}\mathbf{ }\mathbf{two}\mathbf{ }\mathbf{factors}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{intself}\mathbf{.} \\ \mathrm{We} \mathrm{will} \mathrm{assume} 2^{\mathrm{t}}-1 \mathrm{is} \mathrm{prime} \mathrm{and} \mathrm{prove} \mathrm{that} \mathrm{t} \mathrm{is} \mathrm{also} \mathrm{prime}. \\ \mathrm{Now} \\ \mathrm{If} 2^{\mathrm{t}}-1 \mathrm{is} \mathrm{prime} \mathrm{then} \mathrm{t} \mathrm{cannot} \mathrm{be} 1. \\ \mathrm{Now} \mathrm{we} \mathrm{will} \mathrm{show} \mathrm{that} \mathrm{n} \mathrm{cannot} \mathrm{be} \mathrm{composite}. \\ \mathrm{Now} \mathrm{let} \mathrm{t}=\mathrm{ab}, \mathrm{where} \mathrm{a} \mathrm{and} \mathrm{b} \mathrm{greater} \mathrm{than} 1. \\ \Rightarrow 2^{\mathrm{t}}-1=2^{\mathrm{ab}}-1={\left(2^{\mathrm{a}}\right)}^{\mathrm{b}}-1 \\ \mathrm{Let} \mathrm{x}=2^{\mathrm{a}} \\ \Rightarrow 2^{\mathrm{t}}-1={\left(2^{\mathrm{a}}\right)}^{\mathrm{b}}-1={\mathrm{x}}^{\mathrm{b}}-1 \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{x}}^{\mathrm{b}}-1=\left(\mathrm{x}-1\right)\left({\mathrm{x}}^{\mathrm{b}-1}+{\mathrm{x}}^{\mathrm{b}-2}+.....+{\mathrm{x}}^2+\mathrm{x}+1\right)....\left(\mathrm{i}\right) \\ \mathrm{Now} \mathrm{x}=2^{\mathrm{a}}\ge 4 \\ \mathrm{so} \mathrm{x}-1> \\ \mathrm{It} \mathrm{is} \mathrm{easy} \mathrm{to} \mathrm{see} \mathrm{that} \left({\mathrm{x}}^{\mathrm{b}-1}+{\mathrm{x}}^{\mathrm{b}-2}+.....+{\mathrm{x}}^2+\mathrm{x}+1\right)>1 \\ \mathrm{Now} \mathrm{it} \mathrm{follows} \mathrm{from} \left(\mathrm{i}\right) \mathrm{that} {\mathrm{x}}^{\mathrm{b}}-1 \mathrm{i}.\mathrm{e}. 2^{\mathrm{t}}-1 \mathrm{is} \mathrm{the} \mathrm{product} \mathrm{of} \mathrm{two} \mathrm{numbers} \\ \mathrm{that} \mathrm{are} \mathrm{greater} \mathrm{than} 1 \mathrm{which} \mathrm{contradicts} \mathrm{the} \mathrm{fact} \mathrm{that} 2^{\mathrm{t}}-1 \mathrm{is} \mathrm{a} \mathrm{prime} \mathrm{number}. \\ \left(\mathrm{As} \mathrm{prime} \mathrm{number} \mathrm{cannot} \mathrm{be} \mathrm{written} \mathrm{in} \mathrm{product} \mathrm{of} \mathrm{two} \mathrm{numbers} \mathrm{that} \mathrm{are} \mathrm{greater} \mathrm{than} 1 \right) \\ \mathrm{t}=\mathrm{ab} \mathrm{is} \mathrm{not} \mathrm{possible}. \\ \mathrm{Hence} \mathrm{t} \mathrm{must} \mathrm{be} \mathrm{prime}. \\ {2}^{\mathrm{t}}-1 \mathrm{is} \mathrm{prime} \mathrm{then} \mathrm{t} \mathrm{is} \mathrm{prime} \\ \mathrm{or} \mathrm{if} \mathrm{t} \mathrm{is} \mathrm{prime} \mathrm{then} 2^{\mathrm{t}}-1 \mathrm{is} \mathrm{also} \mathrm{prime}. \\ \mathrm{As} \mathrm{P} \mathrm{is} \mathrm{set} \mathrm{of} \mathrm{all} \mathrm{prime} \mathrm{numbers}. \\ \mathbf{Hence}\mathbf{ }\mathbf{S}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{subset}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{P}\mathbf{.} \end{gathered}$$

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