[QUOTE=blob100;215959]If we have (a,N)=g>1,
Where x^2=a(mod N)
(for x,N defined on the last post).
(x,a)=g.
We said that N is a product of K distinct odd primes.
g is devided by k<\=K primes which devide N.
For example:
We give N=3*5*7. K=3.
k<\=3.
We may say that for g devided by K primes which devide N,
We have (K-k)^2 solutions for x.[/QUOTE]

No. You almost have it, but you are still being careless in your
use of variables.

Show the entire derivation.

[quote=R.D. Silverman;215977]No. You almost have it, but you are still being careless in your
use of variables.

Show the entire derivation.[/quote]
For a given a, x varible, N product of K distinct odd primes we have the congruence:
x^2=a(mod N).
For (a,N)=g>1, we have:
g devided by k distinct odd primes which devide N equivallently.
We may say:
(K-k)^2 is the number of solutions for x.

[QUOTE=blob100;216031]For a given a, x varible, N product of K distinct odd primes we have the congruence:
x^2=a(mod N).
For (a,N)=g>1, we have:
g devided by k distinct odd primes which devide N equivallently.
We may say:
(K-k)^2 is the number of solutions for x.[/QUOTE]

Starting that a,x are variables in insufficient. You need to specify
their domain. Be disciplined in what you write.

The answer you give is not correct, and you are not following my
suggestion: Show the [b]entire[/b] derivation.

[quote=R.D. Silverman;216060]Starting that a,x are variables in insufficient. You need to specify
their domain. Be disciplined in what you write.

The answer you give is not correct, and you are not following my
suggestion: Show the [B]entire[/B] derivation.[/quote]
What do you mean by derivation?
The derivation of what?

[QUOTE=blob100;216081]What do you mean by derivation?
The derivation of what?[/QUOTE]

The derivation of your answer. Show all of the steps. Don't just
state the answer.

[quote=R.D. Silverman;216084]The derivation of your answer. Show all of the steps. Don't just
state the answer.[/quote]
I understand,

For the congruence:
x^2=a(mod N)
Where N is a product of K distinct odd primes.
x variable,
(a,N)=g>1.
g is devided by k primes which devide N equivallently.
k<\=K.

We stated before that for (a,N)=1.
We have K^2 solutions for x.
But by assuming a must be devided by k primes (defined before),
We must assume (x,g)=g.
So: (if to go back to the combinations between the solutions), the solutions must go lower.
We see that if we have:
y^2=a(mod q), where y is a variable and q is a divisor of N,
We may say that if a is devided by q, y must be devided by q too,
Which makes the whole congruence and it's solutions useless (this gives x devided by q too).
So, for how many numbers q (we said k) we have (K-k)^2 (number of combinations).

I'm sorry for not explaining the theme well.

[QUOTE=blob100;216100]I understand,

For the congruence:
x^2=a(mod N)
Where N is a product of K distinct odd primes.
x variable,
(a,N)=g>1.
g is devided by k primes which devide N equivallently.
k<\=K.

We stated before that for (a,N)=1.
We have K^2 solutions for x.
But by assuming a must be devided by k primes (defined before),
We must assume (x,g)=g.
So: (if to go back to the combinations between the solutions), the solutions must go lower.
We see that if we have:
y^2=a(mod q), where y is a variable and q is a divisor of N,
We may say that if a is devided by q, y must be devided by q too,
Which makes the whole congruence and it's solutions useless (this gives x devided by q too).

[/QUOTE]

This last assertion is not correct. The congruence is not useless.

[QUOTE]
So, for how many numbers q (we said k) we have (K-k)^2 (number of combinations).

I'm sorry for not explaining the theme well.[/QUOTE]

You still have not explained how you got this final answer.
Explain how you got from the (incorrect) statement that the congruence
is useless, to the very next statement in which you say (incorrectly)
that the number of solutions is (K-k)^2. How do you get "(K-k)^2"?????

[quote=R.D. Silverman;216102]This last assertion is not correct. The congruence is not useless.



You still have not explained how you got this final answer.
Explain how you got from the (incorrect) statement that the congruence
is useless, to the very next statement in which you say (incorrectly)
that the number of solutions is (K-k)^2. How do you get "(K-k)^2"?????[/quote]

OK,
First, to get 2^K (for the last problem where (a,N)=1 was assumed),
We made k+1 congurences where the last one was the modulo N congruence.
And by the combinations we got 2^K.

For the new problem we will start with N=pq for p,q two distinct odd primes.
x^2=a(mod p)
y^2=a(mod q)
z^2=a(mod N)
x,y,z varibles, (a,N)>1.
We may say a=p or q or N.
We may say a=p,
Then p|x,z.
Now, we will go back to the "old" problem,
We needed to find a solution z which "based" on x,y solutions.
x1y1
x2y1
x2y2
x1y2
We may say that for x,y which agree the congruent we have z solution.
Now, by saying x "always" agrees the congruence, we just need to find y which agrees q's congruence and then we may say z agrees too.
We get easily 2^(K-k)...

I'm again sorry for failing writing right.
I always make the same mistakes (writing (K-k)^2 instead of 2^(K-k))
It is just becuase I write fast...
I meant 2^(K-k) instead of (K-k)^2.

[QUOTE=blob100;216214]OK,
First, to get 2^K (for the last problem where (a,N)=1 was assumed),
We made k+1 congurences where the last one was the modulo N congruence.
And by the combinations we got 2^K.

For the new problem we will start with N=pq for p,q two distinct odd primes.
x^2=a(mod p)
y^2=a(mod q)
z^2=a(mod N)
x,y,z varibles, (a,N)>1.
We may say a=p or q or N.
We may say a=p,
Then p|x,z.
Now, we will go back to the "old" problem,
We needed to find a solution z which "based" on x,y solutions.
x1y1
x2y1
x2y2
x1y2
We may say that for x,y which agree the congruent we have z solution.
Now, by saying x "always" agrees the congruence, we just need to find y which agrees q's congruence and then we may say z agrees too.
We get easily 2^(K-k)...

I'm again sorry for failing writing right.
I always make the same mistakes (writing (K-k)^2 instead of 2^(K-k))

[/QUOTE]

Then slow down. Be disciplined.

But you still have not explained where you got 2^(K-k). [it is the correct
answer]


There are missing steps.

[quote=R.D. Silverman;216225]Then slow down. Be disciplined.

But you still have not explained where you got 2^(K-k). [it is the correct
answer]


There are missing steps.[/quote]
I agree there are missing steps,
Is it the correct formula?
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.
By the old problem,
We have combinations between solutions of x and solutions of y to get solutions for z.
"Every solution of z is based on a solution of x and one of y".
If we say x is trivially devided by p then the congurence bellow (the first)
Is always true for these x's (what I mean is: If x agrees the terms, it is a triviall solution (always a solution)).
That so, we are not ,as for the old problem, combine x solutions in the combination of x,y to get z.
We have k=1 (defined on the last posts) and 2^(K-k)=2^(2-1)=2 combinations.

I think it can be proved by induction easily for every K and k.

[QUOTE=blob100;216229]I agree there are missing steps,
Is it the correct formula?
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.
By the old problem,
We have combinations between solutions of x and solutions of y to get solutions for z.
"Every solution of z is based on a solution of x and one of y".
If we say x is trivially devided by p then the congurence bellow (the first)
Is always true for these x's (what I mean is: If x agrees the terms, it is a triviall solution (always a solution)).
That so, we are not ,as for the old problem, combine x solutions in the combination of x,y to get z.
We have k=1 (defined on the last posts) and 2^(K-k)=2^(2-1)=2 combinations.

I think it can be proved by induction easily for every K and k.[/QUOTE]

Yes, it can be proven by induction. But it is not necessary.
You just need to justify the answer. You have not done this.