 Posers and Puzzles

1. 03 Mar '05 11:35
prove that there are infinetly many primes of the form 6k+5.
2. 03 Mar '05 12:39
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.
3. 03 Mar '05 21:23
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.
4. 04 Mar '05 06:135 edits
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

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5. 04 Mar '05 09:401 edit
Sorry, I posted before noticing Acolyte's proof.
Anyway, it's quite interesing that all twin primes must be of the form 6n +/- 1
6. 22 Mar '05 23:06
All primes above 3 are either one more or one less than a multiple of 6.
7. 22 Mar '05 23:09
Originally posted by Bowmann
All primes above 3 are either one more or one less than a multiple of 6.
Been there, proved that.
8. 22 Mar '05 23:11
I simply meant that what you say about twin primes would stand to reason.