Primes 2

**eyke** · February 27th 2010, 03:53 PM

prove that there are infinitely many primes by considering the sequence $2^2(^1) +1 , 2^2(^2)+1 , 2^2(^3)+1 ...$

note the parantheses indicates the power of the first power.

**NonCommAlg** · February 27th 2010, 04:44 PM

Originally Posted by eyke

prove that there are infinitely many primes by considering the sequence $2^2(^1) +1 , 2^2(^2)+1 , 2^2(^3)+1 ...$

note the parantheses indicates the power of the first power.

if $n > m \geq 1,$ then $2^{2^n}+1=(2^{2^m})^{2^{n-m}}+1=(2^{2^m} +1 - 1)^{2^{n-m}}+1 \equiv 2 \mod 2^{2^m} + 1.$ so if $d \mid 2^{2^n}+1$ and $d \mid 2^{2^m} + 1,$ then $d \mid 2,$ which implies that $d=1.$

so every two elements of your infinite sequence are coprime and each term has at least one prime factor. thus the number of primes must be infinite.

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