[quote=R.D. Silverman;217275]Stop inventing terminology.
And your statement: "(I called these "real unities" becuase these are real and unities, ) is a misuse of existing terminolgy. 1 and -1 are units of Z.
In fact, they are the only units. Stop inserting extraneous words into
(i.e. the word 'real', as in 'real unities') your discussion because they
make it appear that you are just confused.


Elements of a ring that are equal to their
own inverse do not have to be 1 or -1. Why do you think that they are?
Consider m = (say) 24. 5 is its own inverse, yet 5 certainly is not 1 or -1.



For about the (3rd? 4th? 5th?) time: PAIR THE UNITS WITH THEIR
INVERSES.




Stop adding words!!!!! (as in 'real' b). Do not use the word real
in a mathematical context unless you are in fact discussing the REAL
NUMBERS. We are discussing the units of the integers taken mod m.
These are the ONLY numbers under discussion. Why do you keep
babbling about "real" numbers? In R, every number [B]except 0[/B] is a unit.

I made it clear at the very beginning of this discussion what an
inverse is. It is a MULTIPLICATIVE INVERSE. How can this possibly be
unclear? The definition of a multiplicative inverse is basic, pre-algebra
arithmetic.[/quote]

The confusion came from not understanding what your talking about.
You said inverse instead of multiplicative inverse.
I'm not confused anymore.

[QUOTE=blob100;217280]The confusion came from not understanding what your talking about.
You said inverse instead of multiplicative inverse.
I'm not confused anymore.[/QUOTE]

My apology. I thought that the word inverse was understood.

I'll show the answer as a criterion:
If we have: (m-1)^2=1(mod m), and there is no g<m-1 prime to m such that: g^2=1(mod m) we have d(m)=-1(mod m).
If we have such a g then d(m)=1(mod m),
Where (m-1)^2=1(mod m) does not needed to exist (it can, but it can be not too).

Examples:
m=24.
We have g={5,7,11,17,19} where g^2=1(mod 24).
d(24)=1(mod 24).

m=10
We have no such a g, but we have (10-1)^2=1(mod 10).
d(10)=-1(mod 10).

I found 24 as a really interesting number.
It has just one quadratic residues which is one itself.
Are there any other numbers of this kind?
Are there infinitely many numbers of this form?

[QUOTE=blob100;217298]I'll show the answer as a criterion:
If we have: (m-1)^2=1(mod m), and there is no g<m-1 prime to m such that: g^2=1(mod m) we have d(m)=-1(mod m).
If we have such a g then d(m)=1(mod m),
Where (m-1)^2=1(mod m) does not needed to exist (it can, but it can be not too).

Examples:
m=24.
We have g={5,7,11,17,19} where g^2=1(mod 24).
d(24)=1(mod 24).

m=10
We have no such a g, but we have (10-1)^2=1(mod 10).
d(10)=-1(mod 10).

I found 24 as a really interesting number.
It has just one quadratic residues which is one itself.
Are there any other numbers of this kind?
Are there infinitely many numbers of this form?[/QUOTE]

Yes, and Yes.

You need to answer the following: What do you get when you multiply
two units, each of which is its own inverse.?

What determines when the unit product is 1 and when it is -1.?

[quote=R.D. Silverman;217300]Yes, and Yes.

You need to answer the following: What do you get when you multiply
two units, each of which is its own inverse.?

What determines when the unit product is 1 and when it is -1.?[/quote]
For what the "Yes and Yes" claiming was directet?

[quote=R.D. Silverman;217300]

You need to answer the following: What do you get when you multiply
two units, each of which is its own inverse.?
[/quote]

When you say so, you mean:
g1g2=b(mod m)
We get b.
And g1 and g2 each units of m with inverse g1 and g2 respectively.
Example:
m=24, g1=5 g2=7.
g1g2=11(mod 24).
b=11.
What happens if there are no two units agree the terms (have thier own inverse)?
I think the product of two units (that agree the terms) modulo m is another unit of that kind.
We may call it a triangle:
For m=16.
15, 9, 7 agree the terms.
15*7=9(mod 16)
15*9=7(mod 16)
7*9=15(mod 16)

Isn't my criterion true?
I'll repeat:
a=1 if and only if there is no such g: g<(m-1) (g,m)=1
g^2=1(mod m), (but m-1 can agree the congruence (m-1)^2=1(mod m)).

a=-1 if and only if there is such g, (m-1 can agree th congruence here too).

[QUOTE=blob100;217335]For what the "Yes and Yes" claiming was directet?[/QUOTE]

To the two questions that immediately preceded the answer

[QUOTE=blob100;217336]When you say so, you mean:
g1g2=b(mod m)
We get b.
And g1 and g2 each units of m with inverse g1 and g2 respectively.
Example:
m=24, g1=5 g2=7.
g1g2=11(mod 24).
b=11.
.[/QUOTE]

Despite my repeated admonshments you are still using some bad habits.

g1, g2, and b are undefined. Stop using variables without defining them!

Stop doing everything by numerical example, and start doing things
algebraically. Numerical examples only HIDE what is going on.

Working mod m:
Let g1 and g2 be units that are their own inverse. Hence,

g1 * g1 = 1 ; Thus elements that are their own inverse have order 2!

g2 * g2 = 1

Whence (g1 * g2)^2 = 1, Thus the product of two units that are
their own inverse is ALSO a unit that is its own inverse.

The purpose of all of these exercizes is for you to discover relationships
among variables that occur in modular arithmetic. STOP doing the numerical
examples and START doing things algebraically.

Now ask yourself: what happens if I multiply ALL the elements of order 2?

[quote=R.D. Silverman;217353]Despite my repeated admonshments you are still using some bad habits.

g1, g2, and b are undefined. Stop using variables without defining them!

Stop doing everything by numerical example, and start doing things
algebraically. Numerical examples only HIDE what is going on.

Working mod m:
Let g1 and g2 be units that are their own inverse. Hence,

g1 * g1 = 1 ; Thus elements that are their own inverse have order 2!

g2 * g2 = 1

Whence (g1 * g2)^2 = 1, Thus the product of two units that are
their own inverse is ALSO a unit that is its own inverse.

The purpose of all of these exercizes is for you to discover relationships
among variables that occur in modular arithmetic. STOP doing the numerical
examples and START doing things algebraically.

Now ask yourself: what happens if I multiply ALL the elements of order 2?[/quote]
First of all, g1 and g2 were defined "each one is a unit of m with inverse g1 and g2 respectivly (g1's inverse is g1 and same with g2)"

The multiply of all the elements of order 2 modulo m is +/-1.
Look: you said: "Whence (g1 * g2)^2 = 1, Thus the product of two units that are
their own inverse is ALSO a unit that is its own inverse."
And I said: "What happens if there are no two units agree the terms (have thier own inverse)?
I think the product of two units (that agree the terms) modulo m is another unit of that kind."
Which is the same..
When I sain "agree the terms" I meant a unit which is it's own inverse.

[QUOTE=blob100;217371]First of all, g1 and g2 were defined "each one is a unit of m with inverse g1 and g2 respectivly (g1's inverse is g1 and same with g2)"
[/QUOTE]

This is not what you wrote. You wrote:

"When you say so, you mean:
g1g2=b(mod m)
We get b."

I could not find a definition of g1 or g2 in your prior posts either.
[QUOTE]

The multiply of all the elements of order 2 modulo m is +/-1.
Look: you said: "Whence (g1 * g2)^2 = 1, Thus the product of two units that are
their own inverse is ALSO a unit that is its own inverse."
And I said: "What happens if there are no two units agree the terms (have thier own inverse)?
[/QUOTE]

For the last time, please stop inventing terminology (i.e. your expression
"agree the terms"). It is clear that "have their own inverse" is well
defined, so why invent terminology? It only adds confusion.

[QUOTE]
I think the product of two units (that agree the terms) modulo m is another unit of that kind."
Which is the same..
When I sain "agree the terms" I meant a unit which is it's own inverse.[/QUOTE]

With regard to this last comment of yours: STOP! Don't say "agree the
terms" when you mean "unit equal to its own inverse".

Try proving the following:

(working mod m)
If x1 and x2 are units with order e1, prove that (x1*x2) has order e1.

Note that if x1 has order e1 and x2 has order e2, then the order of
(x1 * x2) is unpredictable.

So, let us return to the question: What determines when the product of
the units is 1 and what determines when it is -1.??

For this question you may restrict m to be squarefree. (otherwise things
can get complicated because of Hensel's Lemma [which we have not yet
discussed]).

just for laugh...
Euler power hippies?