BdMO Online Forum

Prove that 5th Fermat's number $2^{2^5}+1$ is a multiple of 641.

i tried to write 2^2^5 as dollar($){2^(2^5)+1}dollar($) but didn't worked!!!!!!!!!!someone help me to write like that......
about problem:i used modular and after about restless 8 days i made the solution of it.........now after yours solution I'll post my one............it was the hardest problem I'd ever solved..........

You have towrite 2^{2^5}+1 (I think you made mistake last time)
I mean when you want to write something bigger than 1 letter you have to use second bracket.
So, 2^{2^5}+1=$2^{2^5}+1$

This is due to Euler.
Note that $641=5^4+2^4|2^{32}+5^42^{28}$ and $641=5.2^7+1|5^42^{28}-1$
So $641$ divides their difference $2^{32}+1$

Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.

Dipan wrote:Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.

i dont think you will have access to wikipedia or calculator in any MATH contest .

Dipan wrote:Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.

But this is not allowed in the olympiad and you need to prove this rigorously
And we need the proof,you just checked this.

Sorry..