[quote=axn;217956]What is the product of each pair (mod 16, of course)?
What conclusions can you draw from this observation?[/quote]
1) -1.
The product between a unit of order 2 (denoted as g) and m-g mod m
is -1.
From here, we continue to:
If there are 2k pairs (defined before as: two units of order 2 where thier product is -1 and thier sum is m), the product of [B]all[/B] the units of order 2 mod m is (-1)^k.

[quote=R.D. Silverman;217958]

I don't know which universe you are living in, but in the one I live in,
there are always at least two elements of order 2. Please show
your examples.
[/quote]

It is the same as for 16.
I was not actually including 1 as a unit of order 2 becuase it is of order 1.
We may say there are 4 units of order 2 for 16 (including 1):
1, 7, 9, 15.

An example for a number m from my universe I'm living on is 10 which his units of order 2 are 9 (not including 1), and 2 (including 1).
I guess I need to include 1, but it isn't defined as a number of order 2.

Good night mersenneforum.org.

[QUOTE=blob100;217962]
I guess I need to include 1, but it isn't defined as a number of order 2.[/QUOTE]

No? Elements that are their own multiplicative inverse are elements of
order 2. i.e. x = x^-1 mod m --> x^2 = 1 mod m.

Isn't 1 a solution to x^2 = 1 mod m???

[QUOTE=axn;217940]Don't forget 1 (ok, so technically the order is 1, not 2). The required pairing then would be (1,15), (7,9). What is the pattern here?[/QUOTE]

1 is usually considered to have order 2.

[QUOTE=blob100;217950]Yes, I do see a pattern:
15+1=16
7+9=16.[/QUOTE]

Please, please, stop with the numerical examples!!

Use algebra. The following should have been immediately obvious to anyone
who has taken first year algebra:

If x0 is a solution to x^2 = 1 mod m, then SO IS -x0!!!!
The product : x0 * -x0 = -1!!!!

Pair each element of order 2 with its additive inverse, then multiply the
pairs together. If the number of pairs is odd, then the product is -1.

Why was this so hard?????

[QUOTE=R.D. Silverman;217958]I don't know which universe you are living in, but in the one I live in,
there are always at least two elements of order 2.[/QUOTE]

[QUOTE=R.D. Silverman;217975]No? Elements that are their own multiplicative inverse are elements of
order 2. i.e. x = x^-1 mod m --> x^2 = 1 mod m.

Isn't 1 a solution to x^2 = 1 mod m???[/QUOTE]

[QUOTE=R.D. Silverman;217976]1 is usually considered to have order 2.[/QUOTE]

Okay, now I'm really curious. In a multitude of texts on number theory and algebra, I have [B]never[/B] seen a definition of element order that agrees with this. I have always seen it defined such that the order of the group identity is 1. So when you say it's "usually considered to have order 2", can you point me to the people/texts who are usually doing this considering? Or to put it another way, just which universe [B]are[/B] you living in? :grin:

[QUOTE=jyb;217978]Okay, now I'm really curious. In a multitude of texts on number theory and algebra, I have [B]never[/B] seen a definition of element order that agrees with this. I have always seen it defined such that the order of the group identity is 1. So when you say it's "usually considered to have order 2", can you point me to the people/texts who are usually doing this considering? Or to put it another way, just which universe [B]are[/B] you living in? :grin:[/QUOTE]

You may be (probably/almost certainly are) correct with regard to what
appears in books.

Certainly the common definition of "order" is the smallest n such that a^n = 1,
and for a = 1, n=1 is certainly the smallest.

But I have seen/heard this notational issue debated vigorously during a
West Coast Number Theory Conf. back in the late 80's. One of the
Lenstra's was discussing some open problems related to improving the
efficieny of the (then very new) APR-CL test. It was in the context
not of groups, but of cyclotomic rings. It involved elements of prime
order. Lenstra had tossed a set onto a screen. The set was elements
of order p, but presented in terms of solutions to x^p = 1. Since 1
is a solution, Lenstra had presented it as part of the set of elements of
order p. A debate ensued as to whether this was proper. The general
consensus reached was that in the context of [b]prime power[/b] order,
it was OK to include 1, in order to have to make an exceptional case for
the element 1 when stating a general result. But the consensus was also
that it is highly context sensitive.

I also heard a similar discussion when receiving a lecture on finite fields
from Birkhoff back when I was an undergrad (and yes, we used his book).
I also remember a separate discussion about whether 1 could be considered
as the order of a finite field with 1 element (and indeed whether such a
thing existed) [usually considered the 'trivial' field; whether it is a field
depends on whether one requires the convention that the multiplicative
identity must be different from the additive. ].

Certainly there are many instances when 1 is tossed into a set to keep
from having it singled out as a special case in some theorem. 1 is
excluded from being prime, so we [b]don't[/b] have to create a special
case in the FTA.

I agree that if it were not for the [b]context[/b] of the current problem,
considering 1 to have order 2 would be considered abuse of notation.
I agree that the discussion might have been clearer if we had just stuck
to "elements that are their own inverse". But I deliberately introduced
x^2 = 1, precisely because I was hoping that a discussion of zero divisors
would arise as a result.

I will accept the Lang comment. (Do you know it?) Lang was well known
for telling people at lectures "Your argument is fine, but your notation sucks".

[QUOTE=R.D. Silverman;217992]You may be (probably/almost certainly are) correct with regard to what
appears in books.

Certainly the common definition of "order" is the smallest n such that a^n = 1,
and for a = 1, n=1 is certainly the smallest.

But I have seen/heard this notational issue debated vigorously during a
West Coast Number Theory Conf. back in the late 80's. One of the
Lenstra's was discussing some open problems related to improving the
efficieny of the (then very new) APR-CL test. It was in the context
not of groups, but of cyclotomic rings. It involved elements of prime
order. Lenstra had tossed a set onto a screen. The set was elements
of order p, but presented in terms of solutions to x^p = 1. Since 1
is a solution, Lenstra had presented it as part of the set of elements of
order p. A debate ensued as to whether this was proper. The general
consensus reached was that in the context of [b]prime power[/b] order,
it was OK to include 1, in order to have to make an exceptional case for
the element 1 when stating a general result. But the consensus was also
that it is highly context sensitive.[/QUOTE]

Fair enough, but even then the most you can really say is that the order of 1 will sometimes be considered to be p, for a prime p that is dependent on context. It hardly seems fair to extend that to saying that it is usually considered to be 2, as you did above. And yes, I understand that the context you were interested in here was certainly the case p = 2, but your statement about the "usual meaning" implies that it is context-independent. (Or at least, I certainly took it that way, even though I knew what point you were really trying to make.)

[QUOTE=R.D. Silverman;217992]I also heard a similar discussion when receiving a lecture on finite fields
from Birkhoff back when I was an undergrad (and yes, we used his book).
I also remember a separate discussion about whether 1 could be considered
as the order of a finite field with 1 element (and indeed whether such a
thing existed) [usually considered the 'trivial' field; whether it is a field
depends on whether one requires the convention that the multiplicative
identity must be different from the additive. ]. [/QUOTE]

As an aside, here's perhaps another case where I would balk at your description. I have [B]always[/B] seen fields defined as having 1 != 0. Certainly the trivial [B]ring[/B] is well-defined and useful, but fields always have separate additive and multiplicative identities. And this isn't just arbitrary, it's an important point. For example, there's a theorem that R/I is a field iff I is a maximal ideal of R (R an arbitrary ring). But that only holds if fields are required to have differing identities.


[QUOTE=R.D. Silverman;217992]Certainly there are many instances when 1 is tossed into a set to keep
from having it singled out as a special case in some theorem. 1 is
excluded from being prime, so we [b]don't[/b] have to create a special
case in the FTA.

I agree that if it were not for the [b]context[/b] of the current problem,
considering 1 to have order 2 would be considered abuse of notation.
I agree that the discussion might have been clearer if we had just stuck
to "elements that are their own inverse". But I deliberately introduced
x^2 = 1, precisely because I was hoping that a discussion of zero divisors
would arise as a result.

I will accept the Lang comment. (Do you know it?) Lang was well known
for telling people at lectures "Your argument is fine, but your notation sucks".[/QUOTE]

I wasn't familiar with the Lang comment, but I get the point. In particular, in this case the question of the order of the elements wasn't really of interest. What was much more important was simply whether or not an element was its own inverse (as you say above). And if your audience here had been a bunch of people who could be expected to get that distinction, then it wouldn't much matter.

But your audience was a novice who is trying to learn this material, and when asked for an example of an m for which there are an odd number of elements with order 2, he gave you one, and he was correct, according to the precise definition (the letter of the law, if not the spirit which you intended). And your response was "NO! You should know the definition of element of order 2 by now." But if he had diligently looked up that definition [B]anywhere[/B], he would have reason to think he was correct. In light of that, doesn't your rebuke of him seem a little unwarranted?

All of which, I guess, is to say don't be so quick to give up on Tomer. (I know, I know; you're thinking that the 500+ posts so far make this anything but "quick". :grin:) But he's trying to learn, and it's worth recognizing that not all of his difficulties have been entirely his fault.

You, and others here asked me to give this "numerical example",
I gave it and showed the structure I found.
Further more, I generalized my claim:
The number of pairs (pairs of units of order 2 that so the product of each one is -1) is 2k and the product of the whole pairs mod m is (-1)^k.
I said it (on post 552) and you left it as it is.

[quote=blob100;218001]You, and others here asked me to give this "numerical example",
I gave it and showed the structure I found.
Further more, I generalized my claim:
The number of pairs (pairs of units of order 2 that so the product of each one is -1) is 2k and the product of the whole pairs mod m is (-1)^k.
I said it (on post 552) and you left it as it is.[/quote]

It is easy to prove:
a will define the units of order 2 (a1, a2, a3,..., a2k will be simillarily defined).
There are k pairs of units a{a1,a2,a3,...a2k} ai and aj such that
aiaj=-1(mod m).
We may say that the product of the whole such pairs defined before is
(-1)^k. If k is odd we may say the residue is -1 and if it is even the residue is +1.