there are an infinite number of primes
If with , then clearly and thus . In fact, the order (http://planetmath.org/OrderGroup) of is precisely , for if it were not, say for , then would be a root of and thus would have multiple roots , which is a contradiction. But then, by Fermat’s little theorem, we have , so since is the least integer with this property, we have so that .
We have thus shown that if and , then . The result then follows from the following claim: if is any polynomial of degree at least one, then the factorizations of
contain infinitely many primes. The proof is to Euclid’s proof of the infinitude of primes. Assume not, and let be all of the primes. Since is nonconstant, choose with . Then is clearly divisible by , so , and for each . is nonconstant, so choose such that . Then is clearly divisible by some prime other that the and thus is as well. Contradiction.
Thus the set contains an infinite number of primes in their factorizations, only a finite number of which can divide . The remainder must be primes .
|Title||there are an infinite number of primes|
|Date of creation||2013-03-22 17:43:02|
|Last modified on||2013-03-22 17:43:02|
|Last modified by||rm50 (10146)|