Let p be a prime greater than 3.If there are natural numbers a,b such that all of a

^{p-1}-1, b

^{p-1}-1 and (a+b)

^{p-1}-1 are 0(mod p

^{2}),

then p is a 6n+1 prime

**Moderators:** gmalivuk, Moderators General, Prelates

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 a^{p-1}-1, b^{p-1}-1 and (a+b)^{p-1}-1 are 0(mod p^{2}),

then p is a 6n+1 prime

Let p be a prime greater than 3.If there are natural numbers a,b such that all of a

then p is a 6n+1 prime

well i found these

p (a,b)

all in the form 6n-1, interestingly

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

>-) 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(n^{2}) takes on new meaning when trying to find pairs of socks in the morning.

- gmalivuk
- GNU Terry Pratchett
**Posts:**26739**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

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.

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
- Factoids are just Datas that haven't grown up yet
**Posts:**5967**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

<pedantry> All primes? </pedantry>

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

>-) 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(n^{2}) takes on new meaning when trying to find pairs of socks in the morning.

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 p^{2} (that is between kp^{2} and (k+1)p^{2}), you can always find exactly p-1 numbers d such that d^{{p-1}}-1 is 0(mod p^{2})

2.Can the previous statement be generalised? In other words:

Between two succesive multiples of p^{m} (that is between kp^{m} and (k+1)p^{m}), you can always find exactly p-1 numbers d such that d^{{p-1}}-1 is 0(mod p^{m}).

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 a^{p-1}-1, b^{p-1}-1 and (a+b)^{p-1}-1 are 0(mod p^{3}),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}-d^{p}-1 = 0(mod p^{3}) for p=6n-1 ?

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 p

2.Can the previous statement be generalised? In other words:

Between two succesive multiples of p

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 a

Is the previous statement related to the fact that no one has found solutions to the equation (d+1)

- onoresrts63
**Posts:**29**Joined:**Wed Apr 30, 2014 1:08 pm UTC

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

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 p^{2}(that is between kp^{2}and (k+1)p^{2}), you can always find exactly p-1 numbers d such that d^{{p-1}}-1 is 0(mod p^{2})

2.Can the previous statement be generalised? In other words:

Between two succesive multiples of p^{m}(that is between kp^{m}and (k+1)p^{m}), you can always find exactly p-1 numbers d such that d^{{p-1}}-1 is 0(mod p^{m}).

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 a^{p-1}-1, b^{p-1}-1 and (a+b)^{p-1}-1 are 0(mod p^{3}),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}-d^{p}-1 = 0(mod p^{3}) 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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