[quote=R.D. Silverman;216232]Yes, it can be proven by induction. But it is not necessary.
You just need to justify the answer. You have not done this.[/quote]
For N product of K distinct odd primes p1,p2,p3,...,pk.
Let's take the N=pq ,K=2.
x^2=a(mod p)
y^2=a(mod q)
z^2=a(mod pq=N)
Where g=(a,N)>1
Now, we may say g=N or p or q.
Let's say g=p.
Then p|a,x, which gives p|z.
Then so, we can write:
We will write: (z^2)/p=a/p(mod N/p)
Instead of:
z^2=a(mod pq=N)
Which has 2^(K-1) solutions.
Now generalized:
If we have k numbers p that agree p|a, for N product of K distict odd primes.
We have (z^2)/P=a/P(mod N/P)
For P product of the k primes p.
This new congruence gives 2^(K-k) solutions.
Becuase, N has K odd prime devisors where P lowers it by k, (K-k).

[QUOTE=blob100;216244]For N product of K distinct odd primes p1,p2,p3,...,pk.
Let's take the N=pq ,K=2.
x^2=a(mod p)
y^2=a(mod q)
z^2=a(mod pq=N)
Where g=(a,N)>1
Now, we may say g=N or p or q.
Let's say g=p.
Then p|a,x, which gives p|z.
Then so, we can write:
We will write: (z^2)/p=a/p(mod N/p)
Instead of:
z^2=a(mod pq=N)
Which has 2^(K-1) solutions.
Now generalized:
If we have k numbers p that agree p|a, for N product of K distict odd primes.
We have (z^2)/P=a/P(mod N/P)
For P product of the k primes p.
This new congruence gives 2^(K-k) solutions.
Becuase, N has K odd prime devisors where P lowers it by k, (K-k).[/QUOTE]

More or less, OK. But it can be simplified.

Let N = p_1 * p_2 * .... p_K

Suppose we have a congruence

f(x) = 0 mod p_i and that this congruence has r_i solutions.

Then, using the CRT, f(x) = 0 mod N has r_1 * r_2 * r3.... r_K
solutions.

x^2 = a mod p_i has 2 solutions UNLESS p_i divides a, because now
a = 0 mod _p_1 and x^2 = 0 mod p_i has only ONE solution.
This is the key idea.

Thus, r_i = 2 for all primes that do not divide a, and r_i = 1 if p_i | a.

The result follows immediately.

I want to clearly understand the problem with the product of the units.

For m we have phi(m) units.
If m=10,
By the product of the units you mean d(m)=3*7*9?
Is the problem to find what is d(m) for every m?
If so, can you please give me a clue, or a direction to the problem?

[QUOTE=blob100;216685]I want to clearly understand the problem with the product of the units.

For m we have phi(m) units.
If m=10,
By the product of the units you mean d(m)=3*7*9?
[/QUOTE]

Almost. Although it does not affect the answer to this problem,
1 is also (always!) a unit.


[QUOTE]
Is the problem to find what is d(m) for every m?
If so, can you please give me a clue, or a direction to the problem?[/QUOTE]

I already did. Every unit, [b]BY DEFINITION[/b] has a multiplicative
inverse.

[QUOTE=R.D. Silverman;216710]Almost. Although it does not affect the answer to this problem,
1 is also (always!) a unit.




I already did. Every unit, [b]BY DEFINITION[/b] has a multiplicative
inverse.[/QUOTE]

Note that the original problem is to find the product mod m.

[quote=R.D. Silverman;216711]Note that the original problem is to find the product mod m.[/quote]
Do you mean the residue a of d(m)=a(mod m) where d(m) is the product of the whole units?
By the way, I know 1 is also a unit.
If we take 1 as a factor of d(m) too it does not differ.
As for m=6,
d(m)=1*5.
And a=5?
Is that what I'm suppose to search for?

[QUOTE=blob100;216756]Do you mean the residue a of d(m)=a(mod m) where d(m) is the product of the whole units?
By the way, I know 1 is also a unit.
If we take 1 as a factor of d(m) too it does not differ.
As for m=6,
d(m)=1*5.
And a=5?
Is that what I'm suppose to search for?[/QUOTE]

I don't understand why this is unclear. As a function of m, find the
product of the units mod m.

[quote=R.D. Silverman;216767]I don't understand why this is unclear. As a function of m, find the
product of the units mod m.[/quote]
I'll tell you what is unclear:
I can't understand what you mean by:
"find the
product of the units mod m"
Do you mean: the residue a in the congruence d(m)=a(mod m)?
As for 10, we have d(m)=1*3*7*9
And a=9.
Am I needed to find what is a for every m?

[QUOTE=blob100;216788]I'll tell you what is unclear:
I can't understand what you mean by:
"find the
product of the units mod m"
Do you mean: the residue a in the congruence d(m)=a(mod m)?
As for 10, we have d(m)=1*3*7*9

[/QUOTE]

We have some notational confusion.

This last line is not correct. You are being sloppy.
You defined d(m) := a mod m, yet you simply wrote d(m) = 1*3*7*9
without reducing mod m.

[QUOTE]
And a=9.
[/QUOTE]

a does NOT equal 9 as you have defined it. As you defined it,
a is the product of the units in Z.

a = 1*3*7*9 = 189,


whence d(m) = a mod 10 = 189 mod 10 = 9 mod 10.
as you defined d(m).

[QUOTE]
Am I needed to find what is a for every m?[/QUOTE]



"find the product of the units mod m"

What is unclear about this?


The product of the units is well defined.
This product mod m is well defined. This happens to be a
function of m. This too is well defined.


What is this function?

Perhaps others can help? How am I being unclear?

[quote=R.D. Silverman;216916]


"find the product of the units mod m"

What is unclear about this?


The product of the units is well defined.
This product mod m is well defined. This happens to be a
function of m. This too is well defined.


What is this function?

Perhaps others can help? How am I being unclear?[/quote]
I'll try to express myself better.
Is the product of the units is the finction d(m).
Where d(m) is the product of the numbers g<m, (g,m)=1?

The problem for me is understanding the sentence "find the product of the units mod m".
Do you mean (d(m)/m)?
Can you please just give me an example for specified m.
Show me the product of the units for the specified m, and the product mod m.
Then so i'll figure out what your talking about.

By the way (for the last post), I defined d(m)=189 and a as the residue in the congruence d(m)=a(mod 10).
So we have: 189=9(mod 10), a=9.

[QUOTE=blob100;216943]I'll try to express myself better.
Is the product of the units is the finction d(m).
Where d(m) is the product of the numbers g<m, (g,m)=1?

The problem for me is understanding the sentence "find the product of the units mod m".
Do you mean (d(m)/m)?
Can you please just give me an example for specified m.
[/QUOTE]


I [b]gave[/b] an example.
[QUOTE]
Show me the product of the units for the specified m, and the product mod m.
Then so i'll figure out what your talking about.

By the way (for the last post), I defined d(m)=189 and a as the residue in the congruence d(m)=a(mod 10).
So we have: 189=9(mod 10), a=9.[/QUOTE]


OK. So, if this is your notation, you need to express 'a' as a function of
m. a depends on m.