Applying Chinese Remainder Theorem to polynomials

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SUMMARY

The discussion focuses on solving the system of congruences 7x ≡ 11 mod 30 and 9x ≡ 17 mod 25 using the Chinese Remainder Theorem (CRT) and Bezout's theorem. The key step involves eliminating coefficients by calculating the modular multiplicative inverses of 7 and 9 with respect to their respective moduli. This transforms the system into x ≡ 7-1·11 mod 30 and x ≡ 9-1·17 mod 25. Participants are directed to resources for computing these inverses, including the Extended Euclidean algorithm and online calculators.

PREREQUISITES
  • Understanding of the Chinese Remainder Theorem (CRT)
  • Familiarity with Bezout's theorem
  • Knowledge of modular arithmetic
  • Ability to compute modular multiplicative inverses
NEXT STEPS
  • Learn how to compute modular multiplicative inverses using the Extended Euclidean algorithm
  • Study the application of the Chinese Remainder Theorem in solving linear congruences
  • Explore examples of polynomial congruences and their solutions
  • Investigate online tools for modular arithmetic calculations
USEFUL FOR

Students studying number theory, mathematicians interested in modular arithmetic, and anyone looking to solve systems of linear congruences effectively.

stgermaine
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Homework Statement


Find all integers x such that
7x \equiv 11 mod 30 and
9x \equiv 17 mod 25


Homework Equations


I guess the Chinese Remainder theorem and Bezout's theorem would be used here.



The Attempt at a Solution


I can do this if the x-terms didn't have a coefficient. I'd just rewrite the congruences so that x - 11 = 30k etc and use Euclid's algorithm to solve it, which is not too difficult.
I'm just confused as to what to do since there are the coefficients and I'm not too sure what to do.

Thanks
 
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stgermaine said:

Homework Statement


Find all integers x such that
7x \equiv 11 mod 30 and
9x \equiv 17 mod 25

Homework Equations


I guess the Chinese Remainder theorem and Bezout's theorem would be used here.

The Attempt at a Solution


I can do this if the x-terms didn't have a coefficient. I'd just rewrite the congruences so that x - 11 = 30k etc and use Euclid's algorithm to solve it, which is not too difficult.
I'm just confused as to what to do since there are the coefficients and I'm not too sure what to do.

Thanks

The first step is to get rid of those coefficients by computing their respective modular multiplicative inverses.

So the system becomes ##x \equiv 7^{-1}.11 \pmod {30}## and ##x \equiv 9^{-1}.17 \pmod {25}##.

You can do this because the coefficients are relatively prime to those moduli.

Do you know how to compute those inverses? If not, you can read: http://en.wikipedia.org/wiki/Modular_multiplicative_inverse and http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm (for the actual algorithm). A quick calculator is available online at: http://www.cs.princeton.edu/~dsri/modular-inversion.html
 
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