A number theory problem by mustafa rokadiya

For a natural no. 'n' : 2^{2}^{n}+1 is a prime is : [e.g.: for n=1, 2^{2}^{1} +1 = 5. which is prime no. Hence, it is true.]

true for n=1,2,3,4,5 true for n=1,2 true for n=1,2,3,4 always true

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Josh Speckman
Apr 15, 2014

These numbers are called Fermat Numbers , and for a long time they were thought to be automatically prime. Leonard Euler found that the 5 th 5^{\text{th}} Fermat Number, 4294967297 4294967297 , was not in fact prime. being divisible by 641 641 . Thus the conjecture that all numbers of this form are prime is false, and it is only valid for n = 0 , 1 , 2 , 3 , 4 n={0,1,2,3,4}

Chinmay Patil
Mar 30, 2014

2^2^5 +1= not a prime number

how can you say it is not a prime number. it is 4294967297

Kushagra Sahni - 7 years, 2 months ago

Log in to reply

Try to divide it by basic odd numbers. It's a prime number

Chinmay Patil - 7 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...