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$$x\equiv2\ (\text{mod }6)$$

This one has solution x=2, 8, 14, ...

By multiplying 2 to both sides,

$$2x\equiv4\ (\text{mod }6)$$

By dividing by $2$, $x\equiv 2\ (\text{mod } 3)$ (because $\text{gcd}(2, 6)=2$)

And the solution for this congruence is x=2, 5, 8, ...

What's wrong with this calculation?

Haris Gušić
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won
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1 Answers1

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Note that $$x \equiv y\ (\text{mod }m) \implies ax \equiv ay\ (\text{mod }m)$$ This means that any solution of $x \equiv y\ (\text{mod }m)$ is also a solution of $ax \equiv ay\ (\text{mod }m)$. But, the implication doesn't go both ways, i.e. a solution of $ax \equiv ay\ (\text{mod }m)$ isn't necessarily a solution of $x \equiv y\ (\text{mod }m)$ as well. That is why, in your approach, you got more solutions than the correct answer.

Haris Gušić
  • 4,217
  • Then should I check whether each solutions is correct whenever I get solutions of congruence with multiplying? – won Apr 13 '19 at 08:36
  • @won But, why would you even multiply in the first place? The congruence $x\equiv2\ (\text{mod }6)$ is already the simplest form you can get... You could check each solution individually, but it is easier without even using multiplication the way you did. – Haris Gušić Apr 13 '19 at 08:40
  • But for solving a congruence such as 5x$\equiv$8 (mod 11), I have to multiply 9 to get a solution. So I just wanted to check it doesn't matter to the solution or not. And now I understand why there was a process checking the validity of the solution at the last step in some solutions(If it's wrong, please let me know why) – won Apr 13 '19 at 08:44
  • @won In this case, gcd$(11,9)=1$ so you can multiply the congruence with $9$ and have the same solutions, so you don't have to check each solution. But, for a congruence like $3x\equiv 5\ (\text{mod } 4)$ where gcd$(4,3)\ne 1$, the solutions are not the same when you multiply with $3$. Instead, you should convert this congruence to a linear Diophantine equation and solve it. – Haris Gušić Apr 13 '19 at 08:54
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    Thanks for helping me, now I understand how to solve linear congruence:) – won Apr 13 '19 at 09:22