[QUOTE]I like your humor.[/QUOTE]

I'm glad that you do. :smile:

Now, back to exploring this thread..

[QUOTE]n: natural number.[/QUOTE]

Right..

[QUOTE]e(m)<: the smallest odd element of m.[/QUOTE]

Maths fallacy:M is undefined.

[QUOTE]p: prime element of merssen numbers with an odd exponentiation.[/QUOTE]

This does not follow. O_O

[QUOTE]You can see that 23's smallest odd factor that agrees 2e(p-1)>p is 11 because 23's factors are 2,11.[/QUOTE]

No.. 23 has only one factor: 23.

[QUOTE]83 and 7 are the same as 23.[/QUOTE]

Exactly. 83 and 7 are both primes. 83: 1, 83; 7: 1, 7.

Now, I can make my own trivial conjectures too..

Differences between prime powers:

3^n: 1, 3, 9, 27, 81
Difference between 1 and 3 = 2
Difference between 3 and 9 = 6
Difference between 9 and 27 = 18
Difference between 27 and 81 = 54

Differences follow a pattern of 2 * 3 * 3 * 3 * ..

13^n: 1, 13, 169, 2197, 28561, 371293, ...
Difference between 1 and 13: 12
Difference between 13 and 169 = 156
Difference between 169 and 2197 = 1872

Differences follow a pattern of 12 * 13 * 13 * 13...

In general, for any p^n...
Differences follow a pattern of p-1 * p * p * p * ...

[quote=R.D. Silverman;217137]This is correct. Either a = 1 or a = -1 = m-1 mod m However:

(1) You need to justify your answer. How did you get it?
Show your work. It is not sufficient to simply assert the answer.

(2) You need to determine which m give the answer 1,
and which m give the answer -1.

Hint: This is closely related to a problem we already looked at. Think
about the units that are their own inverse.[/quote]
1)The residue a must be 1 or -1 just becuase it can't be b or c where (b,m)=g>1, 1<(c,d(m))=r<m-1.
a of course not equals b becuase of d(m) will must be devided by b, where d(m) defined as (d(m),m)=1.
a does not equal c becuase:
Lets call d(m)=nf where (d(m),c)=f and n is natural.
nf=fx(mod m) where x is natural, fx=c.
m must be devided by f which is defined as not devided by it.
Then a=+/-1.

[QUOTE=blob100;217234]1)The residue a must be 1 or -1 just becuase it can't be b or c where (b,m)=g>1, 1<(c,d(m))=r<m-1.
[/QUOTE]

Once again, you are being sloppy. You have not DEFINED your
variables. What are b,c, r?

And this statement simply amounts to an assertion that a = 1 or a =-1
simply because it isn't something else. It does nothing to explain WHY.


[QUOTE]
a of course not equals b becuase of d(m) will must be devided by b, where d(m) defined as (d(m),m)=1.

[/QUOTE]

This is gibberish.

[QUOTE]
a does not equal c becuase:
Lets call d(m)=nf where (d(m),c)=f and n is natural.
nf=fx(mod m) where x is natural, fx=c.
m must be devided by f which is defined as not devided by it.
Then a=+/-1.[/QUOTE]

More gibberish.

You have done nothing to explain WHY a = 1 or a =-1 mod m.

[quote=R.D. Silverman;217236]Once again, you are being sloppy. You have not DEFINED your
variables. What are b,c, r?

And this statement simply amounts to an assertion that a = 1 or a =-1
simply because it isn't something else. It does nothing to explain WHY.




This is gibberish.



More gibberish.

You have done nothing to explain WHY a = 1 or a =-1 mod m.[/quote]
I'll try again:
We know (d(m),m)=1.
We will explain why a=1 or a=-1 by explaining a=\=b,c where b is a natural number defined as (b,m)>1.
c defined as a natural number where 1<(c,d(m))<m-1.
We will start proving why a=\=b by contradiction.
We say (m,b)=k.
Let a=b.
d(m)=nk(mod ek) for nk=b, ek=m.
This gives d(m) must be devided by k.
We know that (d(m),m)=1 and on the same time we found d(m) and m devided by k. So, a=\=b.

Now, we will prove a=\=c by contradiction.
We say (d(m),c)=r>1.
nr=er(mod m) for nr=d(m), er=c (e,n are not the same as before, natural numbers which sutisfy the equation but not the last equations nk=b and ek=m).
This gives m must be devided by r.
We know (d(m),m)=1 and on the same time we found d(m) and m devided by r. So, a=\=c.

This gives a must be a real unity.

[QUOTE=blob100;217237]I'll try again:
We know (d(m),m)=1.
[/QUOTE]

You first need to justify this statement. Why is it true?

[QUOTE]
We will explain why a=1 or a=-1 by explaining a=\=b,c where b is a natural number defined as (b,m)>1.
[/QUOTE]

So b divides m.
[QUOTE]
c defined as a natural number where 1<(c,d(m))<m-1.
[/QUOTE]

Why < m-1?

So c divides the product of the units.
[QUOTE]
We will start proving why a=\=b by contradiction.
[/QUOTE]


[QUOTE]

We say (m,b)=k.
Let a=b.
d(m)=nk(mod ek) for nk=b, ek=m.
[/QUOTE]

Huh??? What is e? And the expression for d(m) is
nonsense.

You are using too many variables and doing a poor job defining them.
You have a bad habit of doing this (adding unneeded and undefined
variables into you arguments. They typically just add confusion.)

[QUOTE]
This gives d(m) must be devided by k.
[/QUOTE]


Since k = (m,b) and k > 1, k must divide m. But each unit is co-prime to m.
So where do you get the result that the product of the units is divisible by k?


Your "explanation" is wrong, and does not convey why the product must be
1 or -1. And you *repeatedly* ignore my original hint, which is to [b]PAIR
THE INVERSES[/b]

[quote=R.D. Silverman;217238]You first need to justify this statement. Why is it true?



So b divides m.


Why < m-1?

So c divides the product of the units.





Huh??? What is e? And the expression for d(m) is
nonsense.

You are using too many variables and doing a poor job defining them.
You have a bad habit of doing this (adding unneeded and undefined
variables into you arguments. They typically just add confusion.)




Since k = (m,b) and k > 1, k must divide m. But each unit is co-prime to m.
So where do you get the result that the product of the units is divisible by k?


Your "explanation" is wrong, and does not convey why the product must be
1 or -1. And you *repeatedly* ignore my original hint, which is to [B]PAIR[/B]
[B]THE INVERSES[/B][/quote]
What do you mean by inverse?

[QUOTE=jyb;217145]Sometimes people just make mistakes. It isn't always a sign that they're feeble-minded. For example:
[QUOTE=R.D. Silverman;216971]
e.g. let m = 8. The units are 1,3,5,7. Their product mod 8 is 7. This
clearly isn't 1.
[/QUOTE]
[/QUOTE]

Nice catch! At least someone is paying attention (not me...)

I'll try proving again (I wish less confusing):
m, d(m) and a are defined.
To prove a is a real unity (1 or -1) we will prove it may not be deviding m or d(m).
If a|m, d(m) must be devided by a too, which does not fit it's definition.
If a|d(m), m must be devided by a too, which does not fit d(m)'s definition.
We base it on (d(m),m)=1. If these both (d(m),m) have a devisor a, it does not fit d(m),m)=1.
This gives a must be a real unity.