-1

How do I show that my solution to the congruence is the only solution?

I'm asked to solve, $\mu,\lambda$ that satisfy $$89 \lambda+55 \mu=1$$ Using Euclid's algorithm I found $\lambda=-21,\mu=34$.

Then I'm asked to find the solution to $$89 x \equiv 7\pmod{55}$$ Using $\lambda 89 \equiv1 \pmod{55}$

I found that $7\lambda89 \equiv7 \pmod{55} \implies x=7\lambda +55k$ for $k\in \mathbb{Z}$.

But how do I show, that this is the only solution to the congruence?

gt6989b
  • 54,422
sjm23
  • 419
  • 2
  • 11
  • The solution of $ax+b\equiv 0\pmod m$ is unique if $\gcd(a,m)=1$. – richrow Sep 18 '20 at 20:28
  • So, are $89$ and $55$ coprime? – Geoffrey Trang Sep 18 '20 at 20:31
  • @richrow So you're saying because $gcd(89,55)=1$ there is only one solution?
    Why is that the case? I don't have any theorems saying that. So I think I need to argue for that     – sjm23 Sep 18 '20 at 20:35
  • @GeoffreyTrang yes they are. – sjm23 Sep 18 '20 at 20:36
  • The solution of a general linear congruence $,ax\equiv b\pmod{n},$ is given in the linked dupe thread (as well as many others - see the Linked questions there). If you have specific questions about such then please ask in those threads first, and pose a new question only about the specific points that remain unclear (if you don't get adequate replies there). This helps to improve the quality of prior answers (and localizes information). – Bill Dubuque Sep 18 '20 at 20:48
  • 1
    The uniqueness (and existence) of the solution are explicitly treated in this dupe. If you have further questions please post comments here or there. – Bill Dubuque Sep 18 '20 at 21:15

1 Answers1

0

You found the multiplicative inverse of $89 \bmod {55}$. It's $-21$. Thus we get $89x\cong7\bmod{55}\iff x\cong -21\cdot7\bmod{55}\iff x\cong 18\bmod{55}$.