primes

prove that there are infinetly many primes of the form 6k+5.

Originally posted by genius
prove that there are infinetly many primes of the form 6k+5.

the proff that there are infinite primes is established:
one more or less than any multiple of number b is garenteed not to be a multiple of b.
for any prime p, p!+1 is garenteed to be a larger prime, because p! is a multiple of all numbers p or less, and so p!+1 is not a multiple of any of them. if p!+1 is not prime, then it must have prime multipels greater than p. thus there are infinete primes. p!-1 can also be used.

let us rephrase the term 6k+5 to better fit this argument: 2(3)k-1.
if k is alwas prime, then we have numbers 2(3)k which all have only three prime multipls.

i'll continue this after class.

Originally posted by genius
prove that there are infinetly many primes of the form 6k+5.

Off the top of my head:

Claim: If x is a natural number such that x = -1 mod 6, then x has a prime factor which is congruent to -1 mod 6.

Proof: x is clearly not a multiple of 2 or 3, so its prime factors are all congruent to either 1 or -1 mod 6. But if they were all congruent to 1 mod 6, x would also be congruent to 1 mod 6.

Now suppose that p1,p2,...,pn is a complete list of primes which are congruent to -1 mod 6. Then (p1p2...pn)^2 - 2 is a natural number which
a) is congruent to -1 mod 6
b) doesn't have any of p1,p2,...,pn as a prime factor, contradicting the claim.

Hence there are infinitely many primes of the form 6k+5, QED.


Apparently, it is known that every arithmetic sequence either contains infinitely many primes, or every term in the sequence is divisible by a particular prime. I have no idea how this was proved, but I get the impression that it's quite a hard result.

Originally posted by genius
prove that there are infinetly many primes of the form 6k+5.

Well, all natural numbers are of the form: 6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5

Clearly 6k cannot be prime (divisible by 6), neither can 6k+2 (divisible by 2), 6k+3 (divisible by 3), 6k+4 (divisible by 2).

So the only "prime candidates" are 6k+1 and 6k+5. note that 6k+1 is one after a multiple of 6 and 6k+5 is one before a multiple of 6.

Hence, if we have two consecutive odds that are prime, they will be either side of a multiple of 6.
That is, all twin primes are of the form 6n +/- 1
And the Twin Primes Conjecture remains famously unproven.

http://mathworld.wolfram.com/TwinPrimes.html

.

Sorry, I posted before noticing Acolyte's proof.
Anyway, it's quite interesing that all twin primes must be of the form 6n +/- 1

All primes above 3 are either one more or one less than a multiple of 6.

Originally posted by Bowmann
All primes above 3 are either one more or one less than a multiple of 6.

Been there, proved that.

I simply meant that what you say about twin primes would stand to reason.