Modular Congruence and GCD Proof

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SUMMARY

The discussion centers on proving that if \( a \equiv b \mod n \) and \( d = \gcd(a, n) \), then \( d = \gcd(b, n) \). The proof utilizes the properties of modular arithmetic and the definition of the greatest common divisor (gcd). The key steps involve demonstrating that \( d \) divides both \( b \) and \( n \), and establishing that any common divisor \( c \) of \( b \) and \( n \) must be less than or equal to \( d \). This conclusion is reached by manipulating the equations derived from the modular relationship and the gcd definition.

PREREQUISITES
  • Understanding of modular arithmetic, specifically congruences.
  • Knowledge of the properties of the greatest common divisor (gcd).
  • Familiarity with integer linear combinations and their implications.
  • Basic algebraic manipulation skills.
NEXT STEPS
  • Study the properties of modular arithmetic in-depth, focusing on congruences.
  • Explore the Euclidean algorithm for computing the gcd.
  • Learn about integer linear combinations and their role in number theory.
  • Review proofs involving gcd and modular relationships for deeper understanding.
USEFUL FOR

This discussion is beneficial for students studying number theory, particularly those tackling proofs involving modular arithmetic and the properties of the gcd. It is also useful for educators looking for examples to illustrate these concepts in a classroom setting.

Colleen G
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Homework Statement


If a≡b(mod n) and d=gcd(a.n), prove that d=gcd(b,n).[/B]

Homework Equations


If a≡b(mod n) → n|(a-b) → a-b =nk, for some k∈ℤ → a=nk+b
If d=gcd(a.n) → d=ax+ny
gcd(b,n)=d ↔ d|b and d|n, and if c|b and c|n, then c ≤ a.


The Attempt at a Solution


Since a=nk+b and d=ax+ny, then
d=(nk+b)x+ny
d=nkx+bx+ny
d=bx+n(kx+y)
d=bx+nw, where w=kx+y and w∈ℤ.

I have gotten this far. Originally I thought this was enough to prove d was the gcd of b and n. But now I see that I need to show d|b, d|n, and that if there is a c such that c|b and c|n, then c ≤ d.

Help!
 
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Colleen G said:

Homework Statement


If a≡b(mod n) and d=gcd(a.n), prove that d=gcd(b,n).[/B]

Homework Equations


If a≡b(mod n) → n|(a-b) → a-b =nk, for some k∈ℤ → a=nk+b
If d=gcd(a.n) → d=ax+ny
gcd(b,n)=d ↔ d|b and d|n, and if c|b and c|n, then c ≤ a.


The Attempt at a Solution


Since a=nk+b and d=ax+ny, then
d=(nk+b)x+ny
d=nkx+bx+ny
d=bx+n(kx+y)
d=bx+nw, where w=kx+y and w∈ℤ.

I have gotten this far. Originally I thought this was enough to prove d was the gcd of b and n. But now I see that I need to show d|b, d|n, and that if there is a c such that c|b and c|n, then c ≤ d.

Help!

It's pretty clear that if d|n and d|a then d|b, right? Show that. Now if there is a smaller c that also divides b and n, then it also must divide a and n, right? Show that. Put the two together.
 
Last edited:

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