I'm just trying to study for exam season and got stuck on a question; which I'm not really sure how to tackle.. looking through all my lecture notes I just can't seem to see what techniques are required here.
Let $n\geq 2$ be a positive integer and assume that $p:= 2^{2^n}+1$ is a prime number.
(i) Show that $ p\equiv 2\pmod 5$
I can see $p$ is a Fermat number so I can assume it must be something to do with the properties of them but I just can't seem to be making the connection.. any insight is much appreciated!
Let $p_n=2^{2^n}+1.$
Method 1... $2^4=16\equiv 1 \pmod 5.$ For $n\geq 2$ we have $2^n/4\in \mathbb N$ so $p_n=1+(2^4)^{(2^n/4)} \equiv 1+1^{(2^n/4)}\equiv 1+1 \pmod 5.$
Method 2... $p_2=17\equiv 2 \pmod 5.$ If $p_n\equiv 2 \pmod 5$ then $p_{n+1}=1+(p_n-1)^2\equiv 1+(2-1)^2\equiv 2 \pmod 5.$ By induction on $n\geq 2$ we have it.