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

note the parantheses indicates the power of the first power.

Printable View

- Feb 27th 2010, 03:53 PMeykePrimes 2
prove that there are infinitely many primes by considering the sequence $\displaystyle 2^2(^1) +1 , 2^2(^2)+1 , 2^2(^3)+1 ...$

note the parantheses indicates the power of the first power. - Feb 27th 2010, 04:44 PMNonCommAlg
if $\displaystyle n > m \geq 1,$ then $\displaystyle 2^{2^n}+1=(2^{2^m})^{2^{n-m}}+1=(2^{2^m} +1 - 1)^{2^{n-m}}+1 \equiv 2 \mod 2^{2^m} + 1.$ so if $\displaystyle d \mid 2^{2^n}+1$ and $\displaystyle d \mid 2^{2^m} + 1,$ then $\displaystyle d \mid 2,$ which implies that $\displaystyle d=1.$

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.