Why does 3x ≡ -29 (mod 5) equal to 3x ≡ 1 (mod 5)

Question

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

$$3x ≡ -29 \pmod{5} $$ as

$$3x ≡ 1\pmod{5} $$

Answer 1

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).$

Answer 2

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 $\gcd(n,m)=1$

Answer 3

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:$$3x\equiv 1\mod5$$