Is every Fermat number 2^(2^n) + 1 composite for n > 4?
To determine whether every Fermat number, defined as 2^(2^n) + 1 for n > 4, is composite, we need to examine the properties of Fermat numbers and their factorization.
Fermat numbers are named after the mathematician Pierre de Fermat, who conjectured that all numbers of the form 2^(2^n) + 1 were prime. However, this conjecture was proven incorrect, as the fifth Fermat number (when n = 4) is composite.
To determine if the remaining Fermat numbers for n > 4 are composite, we need to check their factorization. There are several methods to do this, but the most common one is to use primality testing algorithms.
One efficient algorithm is the Miller-Rabin primality test. It is a probabilistic test that quickly determines if a number is composite with a high degree of accuracy. By repeatedly applying this test, we can check if a number is composite or potentially prime.
In the case of Fermat numbers, we can test their compositeness using the Miller-Rabin primality test. If a Fermat number fails this test, we can conclude that it is composite. However, if a Fermat number passes this test for a large number of random bases, it does not necessarily mean that the number is prime. It only means that it passed the test, and further analysis is required.
Applying the Miller-Rabin primality test to each Fermat number for n > 4 will allow us to determine their compositeness. If a number is found to be composite, we can conclude that the statement "every Fermat number 2^(2^n) + 1 is composite for n > 4" is true.