[QUOTE=CRGreathouse;223851]You write p twice and n twice. Is this intentional? That is, do we need to find a pair (n, p) with
[TEX](6n\pm1)p=24n+7[/TEX]
or a quadruple (m, n, p, q) with
[TEX]6np\pm q=24m+7[/TEX]
or a triple (m, n, p) with
[TEX](6n\pm1)p=24m+7[/TEX]
?

My best guess was the last, the literal wording suggests the first, and Pi seems to have guessed the second. OF course other interpretations may be possible.

p and q are primes and m,n natural numbers, presumably.[/QUOTE]

I believe your last is most accurate as well.

[QUOTE=science_man_88]I believe your last is most accurate as well.
[/QUOTE]

Hmm..

[QUOTE=CRGreathouse]My best guess was the last, the literal wording suggests the first, and Pi seems to have guessed the second. OF course other interpretations may be possible.
[/QUOTE]

Well, I based it on his wording: Here's the post:

[QUOTE=science_man_88] I mean when 6n*p+/-p = 24n+7[/QUOTE]

I gave two counterexamples to this, 1567 and 5023.

[QUOTE=science_man_88;223856]I believe your last is most accurate as well.[/QUOTE]

OK, a nice solid mathematical description, some Pari code (posted nicely in a [[i][/i]code] block!)... things are starting to make sense.

[code]for(p=5,30,
for(n=1,20,
if((6*n*p+p)%24==7 || (6*n*p-p)%24==7,
print(floor((6*n*p-p)/24)","p)
)
)
)[/code]

I take it the idea is to replace one or both of the constants 20 and 30 with a large number, then the numbers 6*n*p+p and/or 6*n*p-p give you potential Mersenne primes?

[QUOTE=3.14159;223858]If it's the last one, that's downright impossible.[/QUOTE]

heres my thought originally p is a number >1 I think anyways p*(6n+/-1) will give 2 factors that aren't 1 meaning for p>1 they all show composite so any time they are = 24m+7 by his phrasing they show composite m values I've found patterns to these results for given p as long as we can figure out p*x+c to not name variables again (takes finding a way to find c) then we can find when these hit the sequence given and use any pattern found to eliminate more which will only leave values that generate primes(since 24(n)+7 for the sequence gives odd Mersenne numbers the values that give primes will have to give Mersenne primes).

[QUOTE=3.14159;223858]Well, I based it on his wording: Here's the post:

[QUOTE=science_man_88;223840]I mean when 6n*p+/-p = 24n+7[/QUOTE]

I gave two counterexamples to this, 1567 and 5023.[/QUOTE]

Pi, I'm glad your language skills exceed mine. I can't understand the quoted statement well enough that I could have found a counterexample, and I still don't understand what makes those numbers counterexamples. (I just don't understand the statement, period.) But I'm glad someone here can follow sm88's posts.

[QUOTE=CRGreathouse]I take it the idea is to replace one or both of the constants 20 and 30 with a large number, then the numbers 6*n*p+p and/or 6*n*p-p give you potential Mersenne primes?
[/QUOTE]

But this is impossible, because the result is always divisible by p if he wants both p to be the same. Just try it for yourself.

It's like trying to look for primes of the form 14k + 7. None exist, and none exist here either.

[QUOTE=CRGreathouse]Pi, I'm glad your language skills exceed mine. I can't understand the quoted statement well enough that I could have found a counterexample, and I still don't understand what makes those numbers counterexamples. (I just don't understand the statement, period.) But I'm glad someone here can follow sm88's posts.
[/QUOTE]

I saw it as him stating that 6 * n * a prime ± another prime = 24 * n + 7 is always composite. I posted 1567 and 5023.
But now that he clarified and demanded that both p be the same, there will be no primes found there, because an and b in this case share a common divisor, p.

[QUOTE=3.14159;223862]But this is impossible, because the result is always divisible by p if he wants both p to be the same. Just try it for yourself.

It's like trying to look for primes of the form 14k + 7. None exist, and none exist here either.



I saw it as him stating that 6 * n * a prime ± another prime = 24 * n + 7 is always composite. I posted 1567 and 5023.
But now that he clarified and demanded that both p be the same, there will be no primes found there.[/QUOTE]

yeah I'm an idiot I just have no idea how to put it in a way anyone on here will understand i have tons of ideas but nobody can make sense of them on here so I just hope they can.

[QUOTE=science_man_88]yeah I'm an idiot I just have no idea how to put it in a way anyone on here will understand i have tons of ideas but nobody can make sense of them on here so I just hope they can.
[/QUOTE]

Whenever working on a conjecture, make sure you can't easily prove the conjecture. But, first, refer to the conjecture as an open question until it is answered by either showing it is indeed an open question, easily disproven, or easily proven.

[QUOTE=3.14159;223864]Whenever working on a conjecture, make sure you can't easily prove the conjecture. But, first, refer to the conjecture as an open question until it is answered by either showing it is indeed an open question, easily disproven, or easily proven.[/QUOTE]


I could easily prove it to me just can't express it socially (aspergers suck if this what life will be lol)

[QUOTE=science_man_88;223860]heres my thought originally p is a number >1 I think anyways p*(6n+/-1) will give 2 factors that aren't 1 meaning for p>1 they all show composite so any time they are = 24m+7[/QUOTE]

OK. So my interpretation #3 was somewhat close; what you really meant was
[TEX](6n\pm1)p=24m+7[/TEX] with m,n,p positive integers and p > 1.

Since the right side cannot be divisible by 2 or 3, the factors on the left must be at least 5. Every number > 1 relatively prime to 6 (as the numbers on the left must be) is of the form [TEX]6k\pm1[/TEX], so all the equation says is that you have a composite number that is 7 mod 24. Phew! Why didn't you just say so? :smile:

[QUOTE=science_man_88;223860]by his phrasing they show composite m values I've found patterns to these results for given p[/QUOTE]

Given p... that's good. Now we're looking at multiples of a given p > 3 that are 7 mod 24.

[QUOTE=science_man_88;223860]as long as we can figure out p*x+c to not name variables again (takes finding a way to find c)[/QUOTE]

OK, more variables. I know what p is: we're given it, and we also know that it's relatively prime with 6 and greater than 1. But what are x and c, and what is the significance of [TEX]px+c[/TEX]? Is that supposed to equal something else we know?

[QUOTE=science_man_88;223860]then we can find when these hit the sequence given[/QUOTE]

What sequence?

[QUOTE=science_man_88;223860]use any pattern found to eliminate more[/QUOTE]

At what point have we eliminated anything?

[QUOTE=science_man_88;223860]which will only leave values that generate primes(since 24(n)+7 for the sequence gives odd Mersenne numbers the values that give primes will have to give Mersenne primes).[/QUOTE]

So far my understanding is that you want to consider numbers that are 7 mod 24 and remove the ones that have a factor (other than the number itself, presumably) of the form [TEX]6n\pm1[/TEX]. The method of removing these composite 24m + 7 is based on a sequence or sequences unknown.

[QUOTE=3.14159;223862]But this is impossible, because the result is always divisible by p if he wants both p to be the same. Just try it for yourself.[/QUOTE]

I'm trying to figure out what he means, whether the thing he has in mind is possible or impossible. I'm well able to factor simple expressions, thank you. I'm much less able to interpret what posters here intend. (And not just science_man_88, either, though he's probably one of the hardest.)