03 Mar '05 11:35

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

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my head03 Mar '05 12:39

the proff that there are infinite primes is established:*Originally posted by genius***prove that there are infinetly many primes of the form 6k+5.**

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.

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Loughborough03 Mar '05 21:23

Off the top of my head:*Originally posted by genius***prove that there are infinetly many primes of the form 6k+5.**

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.- Joined
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04 Mar '05 06:135 edits

Well, all natural numbers are of the form: 6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5*Originally posted by genius***prove that there are infinetly many primes of the form 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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