hypothesis on primes

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filipmark
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hypothesis on primes

Postby filipmark » Sun May 04, 2014 11:55 am UTC

Does anyone know if the following statement has been proved or disproved?
Let p be a prime greater than 3.If there are natural numbers a,b such that all of ap-1-1, bp-1-1 and (a+b)p-1-1 are 0(mod p2),
then p is a 6n+1 prime

>-)
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Re: hypothesis on primes

Postby >-) » Sun May 04, 2014 5:28 pm UTC

well i found these

p (a,b)

Code: Select all

59 (506, 53)
83 (161, 99)
227 (1946, 1745)
419 (1841, 1468)
857 (8035, 6810)
971 (8395, 2659)
1109 (7433, 2632)
1223 (5067, 3594)
1259 (10645, 1451)
1283 (11831, 7264)
1289 (9991, 7436)
1493 (10303, 8390)
1637 (8212, 4259)
1847 (15790, 7850)
2423 (21223, 15611)
3191 (7881, 738)
3527 (18766, 5643)
3917 (30168, 22504)
3929 (29765, 23323)
4001 (39047, 10601)
4889 (24460, 17627)
6947 (44579, 22565)
8627 (84271, 63769)


all in the form 6n-1, interestingly

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somehow
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Re: hypothesis on primes

Postby somehow » Mon May 05, 2014 3:14 am UTC

>-) wrote:well i found these

p (a,b)

[snip]

all in the form 6n-1, interestingly


A prime can't be of the form 6n + 0, 6n + 2, or 6n + 4, since it would then be even (with the exception of the one even prime, 2 = 6*0 + 2, obviously).

A prime can't be of the form 6n + 3, since it would then be divisible by three (with the exception of the one divisible-by-three prime, 3 = 6*0 + 3).

Thus any prime (besides 2 and 3) must either be of the form 6n + 1 or 6n + 5.
Xenomortis wrote:O(n2) takes on new meaning when trying to find pairs of socks in the morning.

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gmalivuk
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Re: hypothesis on primes

Postby gmalivuk » Tue May 06, 2014 11:16 pm UTC

Right, but the OP hypothesized that primes fitting the given condition would be of the form 6n+1, and >-) found several of the form 6n-1.
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>-)
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Re: hypothesis on primes

Postby >-) » Wed May 07, 2014 1:24 am UTC

i think somehow was explaining to me why all the primes that weren't 6n+1 had to be 6n-1 and that it wasn't just a coincidence. thanks!

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jestingrabbit
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Re: hypothesis on primes

Postby jestingrabbit » Wed May 07, 2014 5:04 am UTC

<pedantry> All primes? </pedantry>
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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somehow
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Re: hypothesis on primes

Postby somehow » Wed May 07, 2014 11:17 am UTC

>-) wrote:i think somehow was explaining to me why all the primes that weren't 6n+1 had to be 6n-1 and that it wasn't just a coincidence. thanks!

Indeed. Sorry I didn't make that clearer!
Xenomortis wrote:O(n2) takes on new meaning when trying to find pairs of socks in the morning.

filipmark
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Re: hypothesis on primes

Postby filipmark » Thu May 15, 2014 6:26 am UTC

First,a (late) thank to >-) for his answer
Then,I 'd like to put the following thoughts under discussion by anyone who might care (all numbers are natural):
1.Let p b a prime. Between two succesive multiples of p2 (that is between kp2 and (k+1)p2), you can always find exactly p-1 numbers d such that d{p-1}-1 is 0(mod p2)
2.Can the previous statement be generalised? In other words:
Between two succesive multiples of pm (that is between kpm and (k+1)pm), you can always find exactly p-1 numbers d such that d{p-1}-1 is 0(mod pm).
3.All primes p of the 6n+1 form have an (infinite) number of a and b fulfilling the conditions of the OP. On the contrary, primes of the form 6n-1 seldom have such a and b's.
4.An "upgrade" of the OP.
Let p be a prime greater than 3.If there are natural numbers a,b such that all of ap-1-1, bp-1-1 and (a+b)p-1-1 are 0(mod p3),then p is a 6n+1 prime
Is the previous statement related to the fact that no one has found solutions to the equation (d+1)p-dp-1 = 0(mod p3) for p=6n-1 ?

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onoresrts63
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Re: hypothesis on primes

Postby onoresrts63 » Fri May 23, 2014 4:24 am UTC

Oh! Cool idea! Didn't know about this.

DavCrav
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Re: hypothesis on primes

Postby DavCrav » Sat May 31, 2014 12:46 pm UTC

filipmark wrote:First,a (late) thank to >-) for his answer
Then,I 'd like to put the following thoughts under discussion by anyone who might care (all numbers are natural):
1.Let p b a prime. Between two succesive multiples of p2 (that is between kp2 and (k+1)p2), you can always find exactly p-1 numbers d such that d{p-1}-1 is 0(mod p2)
2.Can the previous statement be generalised? In other words:
Between two succesive multiples of pm (that is between kpm and (k+1)pm), you can always find exactly p-1 numbers d such that d{p-1}-1 is 0(mod pm).
3.All primes p of the 6n+1 form have an (infinite) number of a and b fulfilling the conditions of the OP. On the contrary, primes of the form 6n-1 seldom have such a and b's.
4.An "upgrade" of the OP.
Let p be a prime greater than 3.If there are natural numbers a,b such that all of ap-1-1, bp-1-1 and (a+b)p-1-1 are 0(mod p3),then p is a 6n+1 prime
Is the previous statement related to the fact that no one has found solutions to the equation (d+1)p-dp-1 = 0(mod p3) for p=6n-1 ?


2. Yes. The unit group of the ring Z/p^nZ has order (p-1)p^{n-1} and is abelian. Hence there are exactly p-1 solutions to x^{p-1}=1 in this group, namely the elements of order dividing p-1.
3. (xp^2+y)^{p-1}=xp^2(STUFF)+y^{p-1}, so if (a,b) work then so do (p^2+a,p^2+b)


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