[QUOTE=blob100;217917]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).
[/QUOTE]

g(m) is the product of all units of order 2. What do you mean when you say
that the lowest value of g(m) is m-1? You have proved [b]nothing[/b] so far
about the value of g(m). So how can you assert that its "lowest value is
m-1" [whatever you mean by that phrase].?

Your expression "11 < a,b < 12" is total nonsense. There is no element
strictly between 11 and 12. Furthermore, nothing you have done
indicates in [b]any[/b] way that the value of g(m) might be bounded.
Yet you [b]assert[/b] without proof that it is.
Furthermore, I told you to stop using numerical examples!

Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0
mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero
divisor. It does not prove that g(m) = 1 or -1. My example showed
a product that was 0, even though neither element was 0. Why is this
hard to understand?

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


Oh, really? Show an example.

Didn't we previously go through an exercize where we showed that the
number of solutions to x^2 = a mod m was a POWER OF TWO????

Did you pull this latest assertion from your ass? (that the number might
not be even).

I am ready to give up on you. You keep making assertions without
justification, you do not take advice about defining your variables,
you do not take advice about not using unnecessary variables, you
keep resorting to a single numerical example to make an assertion
instead of using algebra. AND YOU DO NOT TAKE HINTS.

[QUOTE=axn;217918]Please find one.[/QUOTE]

I'm ready to give up on him. We spent a lot of time and effort in the
previous exercize to show that the number of solutions to x^2 = a mod m
was a power of two, and then he turns around and asserts that the number
of solutions might be odd!!!!

I have tried to be patient, but he does not seem to listen to my advice.

[QUOTE=R.D. Silverman;217920]g(m) is the product of all units of order 2. What do you mean when you say
that the lowest value of g(m) is m-1? You have proved [b]nothing[/b] so far
about the value of g(m). So how can you assert that its "lowest value is
m-1" [whatever you mean by that phrase].?

Your expression "11 < a,b < 12" is total nonsense. There is no element
strictly between 11 and 12. Furthermore, nothing you have done
indicates in [b]any[/b] way that the value of g(m) might be bounded.
Yet you [b]assert[/b] without proof that it is.
Furthermore, I told you to stop using numerical examples!

Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0
mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero
divisor. It does not prove that g(m) = 1 or -1. My example showed
a product that was 0, even though neither element was 0. Why is this
hard to understand?




Oh, really? Show an example.

Didn't we previously go through an exercize where we showed that the
number of solutions to x^2 = a mod m was a POWER OF TWO????

Did you pull this latest assertion from your ass? (that the number might
not be even).

I am ready to give up on you. You keep making assertions without
justification, you do not take advice about defining your variables,
you do not take advice about not using unnecessary variables, you
keep resorting to a single numerical example to make an assertion
instead of using algebra. AND YOU DO NOT TAKE HINTS.[/QUOTE]


Slight correction. In the previous exercize m was only odd. This
makes a slight difference. However, it does not affect the result
that the number of solutions to x^2= a mod m must be even.
This last result can be proved trivially using bonehead secondary school
first year algebra.

[quote=R.D. Silverman;217920]g(m) is the product of all units of order 2. What do you mean when you say
that the lowest value of g(m) is m-1? You have proved [B]nothing[/B] so far
about the value of g(m). So how can you assert that its "lowest value is
m-1" [whatever you mean by that phrase].?

Your expression "11 < a,b < 12" is total nonsense. There is no element
strictly between 11 and 12. Furthermore, nothing you have done
indicates in [B]any[/B] way that the value of g(m) might be bounded.
Yet you [B]assert[/B] without proof that it is.
Furthermore, I told you to stop using numerical examples!

Look: we have (g(m))^2 = 1 mod m, whence (g(m)-1)(g(m) + 1) = 0
mod m. All this means is that g(m)-1 and g(m)+1 might be ANY zero
divisor. It does not prove that g(m) = 1 or -1. My example showed
a product that was 0, even though neither element was 0. Why is this
hard to understand?




Oh, really? Show an example.

Didn't we previously go through an exercize where we showed that the
number of solutions to x^2 = a mod m was a POWER OF TWO????

Did you pull this latest assertion from your ass? (that the number might
not be even).

I am ready to give up on you. You keep making assertions without
justification, you do not take advice about defining your variables,
you do not take advice about not using unnecessary variables, you
keep resorting to a single numerical example to make an assertion
instead of using algebra. AND YOU DO NOT TAKE HINTS.[/quote]

If you wish to give up, your welcome.
The number of units of order 2 for 16 is 3 (which isn't a 2 power):
7,9,15.
When I say: "11 < a,b < 12 (.....)" I realize you are going to read what is written in the (,).
I meant, that for ab=0(mod 12), you gave 3=a and b=4.
I didn't say anything about g(m)'s bounding.
I didn't use any, but ANY numerical examples on the last posts, I STOPPED.
"Did you pull this latest assertion from your ass? (that the number might
not be even)." Is a really polite sentence.
When I say the lowest value of g(m) may be m-1, I mean:
m-1 is always a unit of order 2, and we can find examples for numbers m with m-1 the only unit of order 2, which gives, g(m)'s value
may be >/=m-1,
We may agree m-1 is always a unit of order 2 becuase:
(m-1)^2=1(mod m)
m^2-2m+1-1=0(mod m)
m^2-2m=0(mod m).

I have a question for you:
Why do you teach me? (I'm fully serious).

Tomer.

[QUOTE=blob100;217935]If you wish to give up, your welcome.
The number of units of order 2 for 16 is 3 (which isn't a 2 power):
7,9,15.[/QUOTE]
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=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]
So why won't we say the number of units of order 2 is 2^n-1 (for specified n natural)?

[QUOTE=blob100;217944]So why won't we say the number of units of order 2 is 2^n-1 (for specified n natural)?[/QUOTE]

Because, for the problem under consideration, we're not worried about order. We're only concerned about units that are their own inverses -- and 1 clearly qualifies. I think this whole order-2 discussion has sidetracked the focus of the problem.

[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]
Yes, I do see a pattern:
15+1=16
7+9=16.

[quote=axn;217948]Because, for the problem under consideration, we're not worried about order. We're only concerned about units that are their own inverses -- and 1 clearly qualifies. I think this whole order-2 discussion has sidetracked the focus of the problem.[/quote]
I see.

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

What is the product of each pair (mod 16, of course)?
What conclusions can you draw from this observation?

[QUOTE=blob100;217935]If you wish to give up, your welcome.
The number of units of order 2 for 16 is 3 (which isn't a 2 power):
7,9,15.
[/QUOTE]

NO! You should know the definition of element of order 2 by now.
You are missing an element.

And you STILL haven't learned from the previous exercize, which was:

How many solutions are there to x^2 = a mod m???

And you still assert that there are only 3 solutions for m = 16.

Don't you believe what you have already learned?

[QUOTE]
When I say: "11 < a,b < 12 (.....)" I realize you are going to read what is written in the (,).
[/QUOTE]

As is everyone else reading this thread.

[QUOTE]
I meant, that for ab=0(mod 12), you gave 3=a and b=4.

I didn't say anything about g(m)'s bounding.
[/QUOTE]

You said 11 < a,b<12. This sure looks as if you imposing bounds
on g to me! The question was about a*b = 0 mod m implying that
a = 0 or b = 0. In this case a = g(m) - 1 and b = g(m) + 1.

[QUOTE]
I didn't use any, but ANY numerical examples on the last posts, I STOPPED.
"Did you pull this latest assertion from your ass? (that the number might
not be even)." Is a really polite sentence.
[/QUOTE]

It was [b] intended [/b] as a rebuke.

[QUOTE]
When I say the lowest value of g(m) may be m-1, I mean:
m-1 is always a unit of order 2,
[/QUOTE]

This last statement is clear. But you did not connect it in any way to
the value of g(m).

[QUOTE]
[and we can find examples for numbers m with m-1 the only unit of order 2,
[/QUOTE]

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]

I have a question for you:
Why do you teach me? (I'm fully serious).

Tomer.[/QUOTE]

I'm still hopeful that you might actually learn something.