Problem 3

Prove that the set of integers of the form

2^k-3

(

k=2,3,\ldots

) contains an infinite subset in which every two members are relatively prime.

Step 3 of 5: Carry every earlier prime factor forward

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p\mid A_k,\ k<n\implies p-1\mid a_n\implies p\mid 2^{a_n}-1

Detailed analysis

If $p$ divides some earlier $A_k$ , then at the next construction it contributes a factor $p-1$ to the exponent, and hence $p-1$ divides every later $a_n$ with $n>k$ . Write $a_n=b(p-1)$ . Fermat's little theorem gives $2^{p-1}\equiv1\pmod p$ , so $p\mid2^{a_n}-1=A_n+2$ .