simplifying congruencies

Question

I'm having a bit of a hard time with a number theory theorem. The statement is as follows:

If $ca \equiv cb \, \text{mod}\, n$, then $a \equiv b \, \text{mod}\, \frac{n}{d}$, where $d=\text{gcf}(c,n)$
(from David M Burton's Elementary Number Theory, Theorem 4.3)

Which means that

if $ca \equiv cb \, \text{mod}\, n$, then $c(a-b)=kn \,, k \in \mathbb{Z}$

so $a-b=k \times \frac{n}{c}$.

My understanding is that when $c=d$ then $c$ is the largest number that both $n$ and $c$ can be divided by, so it gives a sort of "simplest form" for the congruency. But it does not really. For example

$20 \times 40 \equiv 20 \times 10 \quad \text{mod}\, 30$
$\text{gcf}(20, 30)=10$
so $2 \times 40 \equiv 2 \times 10 \quad \text{mod}\, 3$

which is true, but we can still go further and say that $80 \equiv 20 \equiv 2 \quad \text{mod}\, 3$

So stating that $80 \equiv 20 \quad \text{mod}\, 3$ doesn't seem to give more information than if I just divide by ANY common factor of $n$ and $c$.

when $c=5$
$20 \times 40 \equiv 20 \times 10 \quad \text{mod}\, 30$
$160 \equiv 40 \quad \text{mod}\, 6$

which yes, can be further simplified id I divide everything by 2. Is that all? Is that what the theorem states? That there can be no further simplification in terms of congruencies if $c=d$ and it's not about giving the least non negative residue necessarily?

I hope my issue makes sense, any help would be appreciated.

(I might be misunderstanding you) Note that you didn't apply the theorem correctly. Since $ c = 20, a = 40, b = 10, n = 30$ giving $ d = \gcd(20, 30) = 10$, we have $ 40 \equiv 10 \pmod{30/10=3}$. $\quad$ Likewise, later on when you say "when $c = 5$, you are still using $c = 20$ in $20 \times 40 \equiv 20 \times 10$? — Calvin Lin, Mar 17 '25 at 14:28
Thank you. In my example $800 \equiv 200 , \text{mod}, 30$. So just meant to say that if I divide by 10, I get $80 \equiv 20 , \text{mod}, 3$. And to me this doesn't seem to be a "simplest form", nor is $160 \equiv 40 , \text{mod}, 6$ (if I only divide by 5) — Kündücs Eszkábál, Mar 17 '25 at 14:37
The point is that we obtain an equivalent "smaller" congruence by cancelling any common factor $,c,$ of the equation and modulus. Taking the greatest common factor yields the biggest such cancellation simplification, e.g. when solving $,ax\equiv b\pmod{n},,$ where necessarily $,c=\gcd(a,n)\mid b,,$ cancelling the greatest common factor $c,$ forces $a/c$ to be coprime to $n/b,,$ so invertible $!\bmod n/c,,$ so we can divide through by it to solve for $x,,$ yielding $,x \equiv (b/c)/(a/c)\pmod{n/c}\ \ $ — Bill Dubuque, Mar 17 '25 at 15:11
In multi-fraction language $\ x\equiv \dfrac{a}b\pmod{!n},\equiv, \dfrac{a/\color{darkorange}c}{b/\color{darkorange}c}\pmod{!n/\color{darkorange}c},\ $ i.e. cancel $,c,\ \rm\color{darkorange}{everywhere}.\ \ $ See the linked dupes for further elaboration. $\ \ $ — Bill Dubuque, Mar 17 '25 at 15:20
The key point is (nontrivial) cancellation makes the modulus smaller, which generally simplifies matters. $\ \ $ — Bill Dubuque, Mar 17 '25 at 15:57
Thank you! This is very helpful. I'll have a look at the link as well. — Kündücs Eszkábál, Mar 17 '25 at 17:48

score 1 · Answer 1 · answered Mar 17 '25 at 14:42

The theorem doesn't promise to give you a simplest form. (That's a good thing, because that promise is not met, as you've observed.) But it does give you another tool in your toolbox, which is why it's given to you.

Consider the case $$ 1 \cdot 6 = 1 \cdot 6 \bmod 3 $$ That tells you that you can "divide through" by $gcd(1,3) = 1$, i.e., do nothing at all. That's not bad, it just means that this particular theorem isn't the tool to use in this case. Of course, if you rewrite just slightly to get $$ 6 \cdot 1 = 6 \cdot 1 \bmod 3 $$ then you can divide through by $3$ to get $$ 2 \cdot 1 = 2 \cdot 1 \bmod 1 $$ and then, using the fact that any two integers are congruent mod 1, get to $$ 0 = 0 \bmod 1 $$ which is pretty much the limit of what you can do.

simplifying congruencies

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