hypothesis on primes

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filipmark
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Joined: Sun May 04, 2014 9:40 am UTC

hypothesis on primes

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

>-)
Posts: 527
Joined: Tue Apr 24, 2012 1:10 am UTC

Re: hypothesis on primes

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

somehow
Posts: 99
Joined: Wed Aug 14, 2013 8:31 pm UTC

Re: hypothesis on primes

>-) 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.

gmalivuk
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Re: hypothesis on primes

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.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
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(he/him/his)

>-)
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Re: hypothesis on primes

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!

jestingrabbit
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Re: hypothesis on primes

<pedantry> All primes? </pedantry>
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somehow
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Re: hypothesis on primes

>-) 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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Joined: Sun May 04, 2014 9:40 am UTC

Re: hypothesis on primes

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.
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 ?

onoresrts63
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Re: hypothesis on primes

DavCrav
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Re: hypothesis on primes

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.