Problem about Fermat numbers and order of an element

[demonelite123]

Let F_n = 2²ⁿ + 1 be the nth Fermat number and suppose that p|F_n, where p is a prime (possibly F_n itself). Show that 2^{2^{n+1}} \equiv 1 (mod p) so that ord_p(2)|2^{n+1}. Use this to show that ord_p(2) = 2^{n+1}.

So far i have shown that 2^{2^{n+1}} \equiv 1 (mod p) so that ord_p(2)|2^{n+1}. But what I'm having trouble showing now is that 2ⁿ⁺¹ is the smallest possible number k such that 2^k \equiv 1 (mod p). I know that the order of 2 must be some power of 2 so I have tried to use a contradiction argument assuming that there exists a 2^k such that 2^{2^k} \equiv 1 (mod p). then i have 2^{2^{n+1}} \equiv 2^{2^k} (mod p) so 2^{2^{n+1}-2^k} \equiv 1 (mod p). but i can't seem to find a contradiction.

can someone give me some hints to continue? thanks.

 

[dodo]

2^{2^{n+1}-2^k} = 2^{\left(2^k + 2^{k+1} + 2^{k+2} + ... + 2^n\right)} = 2^{2^k} \cdot 2^{2^{k+1}} \cdot 2^{2^{k+2}} \cdot ~...~ \cdot 2^{2^n}, but 2^{2^k} was 1 modulo p... See if you can unwind the whole thing to get 2^{2^{k+1}} \equiv 1 \pmod p, at which point you can find your contradiction.

Be careful, however (you probably know this already), with arguments that cancel something on both sides of a congruency. 10 \equiv 6 \pmod 4, but 5 \not\equiv 3 \pmod 4. What you cancel out in your case are expressions congruent to 1 mod p, which, since they are 1 multiplying something, can be removed.

 

[demonelite123]

i think i know how to continue the problem. thanks for your reply!

 

[dodo]

Oops... I used a neat trick to conclude something stupid. Sorry, it was Sunday. :)

Successive squares of 2^{2^k} will also be congruent to 1, which contradicts the initial statement that 2^{2^n} \equiv -1 \pmod p (since p has to be odd to divide F_n).

 

[demonelite123]

ah yes that is what i have concluded as well. your first post still helped me to come to fact though!