[QUOTE=R.D. Silverman;217747]Note that the "last question" was not:
If x1 and x2 are units with order e1, prove that (x1*x2) has order e1.

The "last question" was to discover what determines when the product
is +1 and when it is -1. However, it *can* be done without the heavy
group-theory machinery. I am trying to lead Tomer through that now.[/QUOTE]

Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason?

[QUOTE=jyb;217755]Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason?[/QUOTE]

e1 = 2 here.

[QUOTE=R.D. Silverman;217781][QUOTE=jyb;217755]Ah. Indeed, I misunderstood which was the "last question" there. But then that suggests that you still want Tomer to try to prove the other claim ("If x1 and x2 are units with order e1, prove that (x1*x2) has order e1"), which will naturally be impossible. Or was that your point, for some reason?[/QUOTE]

e1 = 2 here.[/QUOTE]

It's still false. Unless you stipulate that x1 != x2, of course. But even then, why did you generalize the problem from order 2 to order e1 if the point you were making only works for order 2? Particularly since you had already shown it to be true for order 2 in this post: [url]http://mersenneforum.org/showpost.php?p=217353&postcount=503[/url].

I'll back to the tuples:
A tuple g is defined as:
The product of 3 specified distinct units of order 2.
The specified 3 units are:
Given 2 units of order 2 which thier product mod m is the third unit of order 2.
Two tuples can have a common devisor.
Example:
m=24.
5*7=11(mod 24)
With the numbers 5,7,11 we make the tuple 5*7*11.
On the same time we have:
5*13=17(mod 24)
A common devisor 5 is recieved by the two tuples (5*7*11 and 5*13*17).

[QUOTE=jyb;217807]It's still false. Unless you stipulate that x1 != x2, of course. But even then, why did you generalize the problem from order 2 to order e1 if the point you were making only works for order 2? Particularly since you had already shown it to be true for order 2 in this post: [url]http://mersenneforum.org/showpost.php?p=217353&postcount=503[/url].[/QUOTE]

I thought that it had been stipulated that the question was about
the product of two different units of order 2.

If I failed to specify e1 = 2, it was an oversight on my part.

[QUOTE=blob100;217810]I'll back to the tuples:
A tuple g is defined as:
The product of 3 specified distinct units of order 2.
The specified 3 units are:
Given 2 units of order 2 which thier product mod m is the third unit of order 2.
Two tuples can have a common devisor.
[/QUOTE]

Since all elements are units, it is meaningless to speak of a 'divisor'.
Every unit has a multiplicative inverse, so every unit is a divisor of
all other units.

[QUOTE]
Example:
m=24.
5*7=11(mod 24)
With the numbers 5,7,11 we make the tuple 5*7*11.
On the same time we have:
5*13=17(mod 24)
A common devisor 5 is recieved by the two tuples (5*7*11 and 5*13*17).[/QUOTE]

This last statement is pure [b]NONSENSE[/b]. You are working with
units mod m, and [b]any[/b] unit will be a "common divisor" of
5*13*17, because every unit has an inverse! Given any unit x0
then x0 divides 5*7*11 and it divides 5*13*17 mod 24. You are not
working in Z.


I am almost ready to give up on you because:

(1) I told you to forget about 3-tuples of units, yet you persist with them.

(2) I told you to [b]stop[/b] discussing mathematics from numerical
example(s) and to use algebra, yet you still insist on trying to discuss
an algebraic question by presenting numerical examples.



Your problem: prove that the product of (all) units of order 2 is either 1
or -1.

I already told you (and explained why) that your "3-tuple" approach will
not work. Try something else. The proof is only a couple of lines and
is actually quite easy.

[quote=R.D. Silverman;217813]Since all elements are units, it is meaningless to speak of a 'divisor'.
Every unit has a multiplicative inverse, so every unit is a divisor of
all other units.



This last statement is pure [B]NONSENSE[/B]. You are working with
units mod m, and [B]any[/B] unit will be a "common divisor" of
5*13*17, because every unit has an inverse! Given any unit x0
then x0 divides 5*7*11 and it divides 5*13*17 mod 24. You are not
working in Z.


I am almost ready to give up on you because:

(1) I told you to forget about 3-tuples of units, yet you persist with them.

(2) I told you to [B]stop[/B] discussing mathematics from numerical
example(s) and to use algebra, yet you still insist on trying to discuss
an algebraic question by presenting numerical examples.



Your problem: prove that the product of (all) units of order 2 is either 1
or -1.

I already told you (and explained why) that your "3-tuple" approach will
not work. Try something else. The proof is only a couple of lines and
is actually quite easy.[/quote]

First of all, This last post had not titled with "This post is the approach..."
It was meant to explain a missunderstand of of you.
I know what is the problem I'm meant to solve.
Why won't you give me the directon to solve the problem,
I deffinietly don't understand how to attack this problem.
You gave me the hint: (-1)^n.
I tought this may be helpfull myself, the hint does not help me...
Why did you on one of the posts concerned the product of the units of order 2 mod m? This is what gave me the idea of the tuples.

If we have g(m) the product of the whole units of order 2,
We have (g(m))^2=1(mod m).
(g(m)-1)(g(m)+1)=0(mod m)
Which gives m is a devisor of g(m)-1 or g(m)+1,
Which is our claim.

I tought I firstly need to examine whether the residue is 1 or -1,
That is why I used the tuples (I won't talk about them anymore as you wish).

[QUOTE=blob100;217821]If we have g(m) the product of the whole units of order 2,
We have (g(m))^2=1(mod m).
(g(m)-1)(g(m)+1)=0(mod m)
Which gives m is a devisor of g(m)-1 or g(m)+1,
[/QUOTE]


Go to google. Look up "zero divisor".
If a*b = 0 mod m, where m is composite, it need not
be the case that either a or b is zero. e.g. 3*4 = 0 mod 12, but neither
3 not 4 is zero.

I will add a hint. Pair the units of order 2 in an appropriate way, so that
the product of each pair is -1.

[quote=R.D. Silverman;217844]Go to google. Look up "zero divisor".
If a*b = 0 mod m, where m is composite, it need not
be the case that either a or b is zero. e.g. 3*4 = 0 mod 12, but neither
3 not 4 is zero.

I will add a hint. Pair the units of order 2 in an appropriate way, so that
the product of each pair is -1.[/quote]
First of all, your example is not able to be given:
g(m) lowest value is m-1, that so, we may have 11<a,b<12 (where you gave a=3 and b=4).
Secondly:
About the hint: we can find (sometimes) that the number of units of order 2 isn't even.

[QUOTE=blob100;217917]Secondly:
About the hint: we can find (sometimes) that the number of units of order 2 isn't even.[/QUOTE]

Please find one.