Inverse chinese remainder theorem

Click For Summary
SUMMARY

The discussion centers on the Inverse Chinese Remainder Theorem and methods to determine if a number is not a solution to a set of congruences. Participants suggest that while plugging a number into the congruences is a straightforward method, it becomes inefficient with larger numbers or multiple congruences. An analytical approach involving the greatest common divisor (gcd) or least common multiple (LCM) is proposed for efficiency. The Chinese Remainder Theorem guarantees unique solutions under certain conditions, but exceptions exist when moduli share common factors.

PREREQUISITES
  • Understanding of the Chinese Remainder Theorem
  • Knowledge of modular arithmetic
  • Familiarity with concepts of gcd and LCM
  • Basic algebraic manipulation skills
NEXT STEPS
  • Research advanced applications of the Chinese Remainder Theorem in cryptography
  • Learn about algorithms for efficiently solving systems of congruences
  • Explore methods for calculating gcd and LCM in programming
  • Investigate the implications of non-unique solutions in modular systems
USEFUL FOR

Mathematicians, computer scientists, and anyone interested in number theory, particularly those working with modular arithmetic and congruences.

juan avellaneda
Messages
37
Reaction score
0
hi all I am new on the forum

I wonder if is possible to find a method that proofs that a number IS NOT a solution of a set of congruences
Maybe using the chinese remainder theorem??

best regards
japam
 
Physics news on Phys.org
Well, simply plugging in the number in the congruences and seeing it they become true should work, right?
 
method

thats a solution

but i was thinking in other possibility, a mathematical condition
involving the gcd or MCM, for example

im sure that it exists but i don't know how to probe that
 
I can't imagine any way that could be simpler than plugging your number into the equation.

Is there something unusual about your problem that this is not a desirable technique?
 
method

this method is disadvantageous when you have a big number , or a big number of congruences to check

And suppose we want to know all the numbers that don't fit anyone of a set of congruences

In this case, clearly we need a more analytical method
 
this method is disadvantageous when you have a big number , or a big number of congruences to check

I don't see why.



Anyways, it seems that what you really want to do is to spend a lot of time precomputing information about a system of congruences so that, in the future, you can tell quickly if a number satisfies the system.

The easiest way I can imagine to do this is to simply solve the system of congruences, and then your test is simply comparing a given number to the solution.

e.g. if the system was

3x = 4 (mod 5)
x = 7 (mod 11)

The solution is

x = 18 (mod 55)


So for any given integer n, you simply check if n = 18 (mod 55).
 
counterexample

No, i think is wrong

for example

X = 2 (mod 3)

X = 3 (mod 5)

the answer is X = 8 (mod 15)

Now i check the num 11 that doesn't fit the last congruence
but 11 fits the first

so its not a sufficient proof
 
Hurkyl,

This math is new to me. How do you obtain x = 18 (mod 55)?
 
Ok, so you want integers that satisfy none of the congruences.

Again, I cannot divine a better method than simply plugging the integer into each of the congruences.

You could achieve some speed-up, possibly, by computing a divisibility test for each of your moduli. Or, better yet, make a giant array whose length is equal to the LCM of all the moduli, and store "true" and "false" for each entry.


Loren: The problem is small enough that I did it by trial and error. :smile: In general, you can use the Chinese Remainder theorem.
 
  • #10
There is a unique answer.

juan avellaneda said:
hi all I am new on the forum

I wonder if is possible to find a method that proofs that a number IS NOT a solution of a set of congruences
Maybe using the chinese remainder theorem??

best regards
japam

According to the Chinese Remainder Theorem if we have moduli X,Y,Z...which have no common factor, then a specific congruent solution is unique up to multiples of XYZ...

Proof: Let us suppose that W == r, Mod X, W==m Mod Y, W == s Mod Z...etc. And some other number T also satisifies the same set of congruences. If W is not T, consider W-T = A. If we look at the set of congruences we find: W-T = A == 0 Mod X, Y, Z...since W and T have the same residues in that system.

So we find that A is divisible by X, Y, Z, etc...and so A ==0 Mod XYZ...

This means A = KXYZ..., where K is an integer. Thus a solution to the Chinese Remainder problem is unique up to the product of the relatively prime moduli.

PS It might be added that uiqueness would not work in the following situation:

X==0 Mod 2, X==2 Mod 8, X==10 Mod 16. Thus X = 10 works in all three cases, but the product of the modulli is 256, while X=26 also works!
 
Last edited:

Similar threads

MHB Understanding the Chinese Remainder Theorem for $\mathbb{Z}^{\times} _{20}$
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Use of induction in the proof of the Chinese Remainder Theorem
  • · Replies 7 ·
Replies
7
Views
3K
Graduate Generalization of Chinese Remainder Theorem
  • · Replies 1 ·
Replies
1
Views
4K
Why Does Expanding (13-8)^99 Not Yield the Same Remainder as 5^99 Modulo 13?
  • · Replies 7 ·
Replies
7
Views
2K
Graduate Chinese Remainder Theorem, Solving For Multiplicative Inverses
  • · Replies 2 ·
Replies
2
Views
6K
MHB Remainder/factor theorem question
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad ##(p^k -1) \equiv X \mbox{(mod p)}## via Wilson's theorem
  • · Replies 2 ·
Replies
2
Views
1K
Graduate The Chinese Remainder Theorem for moduli that aren't relatively prime
  • · Replies 2 ·
Replies
2
Views
4K
Graduate Is the Proof in Dummit and Foote's Chinese Remainder Theorem Inductive?
  • · Replies 1 ·
Replies
1
Views
4K
Graduate How do I use Chinese remainder theorem to solve for x mod 683 in Cryptography?
  • · Replies 12 ·
Replies
12
Views
4K