Question

show that if \(2^k+1\) is an odd prime then \(k = 2^n\) for some \(n\ge 0\)

Answer 1

Do the following steps:
Step 1: Show it is true for n=1
Step 1: Assume it is true for n=k
Step 2: Prove it is true for n=k
Step 3: Show it is true for n=k+1

Answer 2

Found this on wikipedia - http://en.wikipedia.org/wiki/Fermat_number#Other_theorems_about_Fermat_numbers

Answer 3

induction looks interesting but im not sure if it is easy to induct on n or k because both are arbitrary here

Answer 4

thats a nice proof @BAdhi
for an alternative proof, the number 2^m + 1 in binary looks promising... but im still working on this

Answer 5

(stuck actually)

Answer 6

\[\large
\begin{array}{}
2^{2*3}+1 &= 1000001 &= 101*1101\\
2^{3*3}+1 &= 1000000001 &= 1001*111001\\
2^{4*3}+1 &= 1000000000001 &= 10001*11110001\\
\cdots
\end{array}
\]