I'm having some trouble understanding the following problem, why can you write the following congruence:

$$3x ≡ -29 \pmod{5} $$ as

$$3x ≡ 1\pmod{5} $$

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    $\begingroup$ Because $-29\equiv 1 \pmod 5$. $\endgroup$ – lulu Jun 19 at 13:50
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    $\begingroup$ $3x+5 \equiv -24 \ldots \implies 3x+15 \equiv -14\,\ldots \implies 3x+30 \equiv 1\, \implies 3x \equiv 1\, \quad (\text{Mod}\, 5)$ $\endgroup$ – Kevin Jun 19 at 13:54
  • $\begingroup$ in short just add 30 on both side and know that $30=0 (mod 5)$ $\endgroup$ – JustWandering Jun 19 at 13:57
  • $\begingroup$ Why is this question downvoted? It's a perfectly legitimate one? $\endgroup$ – David Jun 19 at 14:10
  • $\begingroup$ @David Agreed, I upvoted. $\endgroup$ – Kevin Jun 21 at 13:54

We have that $1-(-29)=30$ so $5$ divides to $1-(-29)$. That means that $-29 \equiv 1 \ (\mbox {mod }5)$. By transitivity we have that if $ 3x \equiv -29 \ (\mbox {mod }5)$ and $-29 \equiv 1 \ (\mbox {mod }5)$ then $ 3x \equiv 1 \ (\mbox {mod }5).$

  • $\begingroup$ Thanks for the answer, but how did you get to −29≡1 (mod 5) thats the part i dont really understand $\endgroup$ – kokayy Jun 19 at 14:20
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    $\begingroup$ @kokayy $-29 \equiv 1 (mod 5)$ because $-29 = (-6)*5 +1$ and $1 = (0)*5 +1$ Similarly, 11, 16, -4, -9... are also congruent with 6 $\endgroup$ – David Jun 19 at 14:22
  • $\begingroup$ @David ohh makes perfect sense now! thanks alot $\endgroup$ – kokayy Jun 19 at 14:23
  • $\begingroup$ @kokayy If $m \in \mathbb{N}$ and $a,b \in \mathbb{Z}$ then $a \equiv b \ \mbox{(mod }m)$ if and only if $m$ divides to $b-a$. $\endgroup$ – Mainkit Jun 19 at 15:56

Keep in mind that, in $\mathbb{Z}/5\mathbb{Z}$, 5 and 0 are the same $5 \equiv 0 (\mod 5)$

Since 5 and 0 are the same, you can add $0$ to the left -hand side of the equation and $5$ to the right-hand side, because you are actually adding $0$ on both sides.

This reasoning works for every multiple of $5$, $30$ included. Similarly, you could add $6$ to one side of the equation and $13$ to the other, without altering the solutions.

Also, since $\mathbb{Z}/5\mathbb{Z}$ happens to be a field ($5$ is prime), you can do the same thing for multiplication except, of course, you are not allowed to multiply by $5$ (because it's $0$!), so it's the same rules as with equations in $\mathbb{R}$

If you were in $\mathbb{Z}/n\mathbb{Z}$, with n not prime, you would be able to multiply both sides of the equation by $m$ (without creating any new false solutions) as long as $\gcd(n,m)=1$


Alternative look.

In e.g. the theory of abelian groups

$$3x\equiv -29\mod5$$ expresses exactly that: $$\{3x+5n\mid n\in\mathbb Z\}=\{-29+5n\mid n\in\mathbb Z\}$$

(This especially when we are dealing with group $\mathbb Z/5\mathbb Z$)

Note that here the equivalence sign is replaced by the equality sign, which might make things more simple to grasp.

Now observe that it is not difficult to prove that $\{-29+5n\mid n\in\mathbb Z\}=\{1+5n\mid n\in\mathbb Z\}$ so that the equality can also be written as:$$\{3x+5n\mid n\in\mathbb Z\}=\{1+5n\mid n\in\mathbb Z\}$$or again as:$$3x\equiv 1\mod5$$

  • $\begingroup$ Be aware that congruence is not normally defined to be "actually a notation" for an equality of (co)sets. Rather, the usual definition is that $\,a\equiv b\pmod{\! n}\iff n\mid a-b.\ $ Usually the coset viewpoint is not considered at length until quotient rings are introduced (often much later - perhaps even in a different book/course). $\endgroup$ – Bill Dubuque Jun 19 at 14:45
  • $\begingroup$ @BillDubuque Yes, you are right. I have decided to delete this answer within some minutes. $\endgroup$ – drhab Jun 19 at 14:48
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    $\begingroup$ You could instead just mention it as another viewpoint instead of "actually a notation" and that would help to avoid any confusion. No doubt that it is a useful viewpoint. $\endgroup$ – Bill Dubuque Jun 19 at 14:49
  • $\begingroup$ @BillDubuque So I did then. Thank you for your comment. $\endgroup$ – drhab Jun 19 at 14:56

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