Number Theory Modulo

Question

I need to prove that if $k\cdot a \equiv k\cdot b \ \ (mod \ n)$ then $a \equiv b\left( \mod \ \frac{n}{gcd(k,n)} \right)$

I tried to do this: $$ k(a-b)=s\cdot n \\ k=u\cdot \gcd(k,n) \wedge n=v\cdot \gcd(k,n) \ ; \gcd(u,v)=1 \\ u\cdot \gcd(k,n) \cdot (a-b)=s\cdot v \cdot \gcd(k,n) \\ u\cdot (a-b)=s\cdot v $$ and now I'm stuck, If anyone has some idea on how can I continue from this step or is there another way to start, I will be grateful!

score 1 · Answer 1 · answered Jun 20 '21 at 15:16

1

Now you have that $ua \equiv ub \pmod {v}.$ But $\gcd(u, v) = 1,$ and so $u^{-1}$ exists mod $v$. Multiply by it to get $a \equiv b \pmod{v}.$

answered Jun 20 '21 at 15:16

How do I know $u^{-1}$ exists? – JollyQ1 Jun 20 '21 at 15:31
@JollyQ1 Because $\gcd(u, v) = 1$ – Jun 20 '21 at 15:39
yes but how do i prove it? – JollyQ1 Jun 20 '21 at 15:45
@JollyQ1 What number theory do you know? This is a standard but tricky lemma, usually proven with the Euclidean algorithm. – Jun 20 '21 at 17:57
Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 20 '21 at 18:20

Number Theory Modulo

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