# $a\equiv b\pmod{n}\iff a/x\equiv b/x\pmod{n/\gcd(x,n)}$ for integers $a,b,x~(x\neq 0)$ and $n\in\Bbb Z^+$?

I'm trying to prove/disprove the following:

If $a,b,x$ be three integers (where $x\neq 0$) such that $x\mid a,b$ and $n$ be a positive integer, then the following congruence holds:

$$a\equiv b\pmod{n}\iff a/x\equiv b/x\pmod{n/\gcd(x,n)}$$

My intuition says it's true and below here is my attempt at a proof. Can the community verify if it's correct? Thanks.

Proof.

Necessity: If $a/x\equiv b/x\pmod{n/\gcd(x,n)}$, then we can write $(a-b)/x=nl/\gcd(x,n)$ for some integer $l$, so that we have,

$$(a-b)/n = lx/\gcd(x,n) = l\ast (x/\gcd(x,n))$$

Since $\gcd(x,n)\mid x$ by definition of $\gcd$, we see that $(a-b)/n$ is an integer and hence $a\equiv b\pmod n$

Sufficiency: If $a\equiv b\pmod n$, then we can write $a-b=np$ for some integer $p$, so that we have,

\begin{align}a-b=np&\implies (a-b)\gcd(x,n)=np\gcd(x,n)\\&\implies \frac{(a-b)\gcd(x,n)}{nx}=\frac{p\gcd(x,n)}{x}\end{align}

By Bezout's Lemma, there exists integers $b_1,b_2$ such that $\gcd(x,n)=b_1x+b_2n$, so we have,

$$\frac{(a-b)\gcd(x,n)}{nx}=\frac{p(b_1x+b_2n)}{x}=pb_1+\frac{b_2np}{x}=pb_1+b_2\frac{a-b}{x}$$

Since $x\mid a,b$, we have $x\mid a-b$ and so we can write $a-b=xq$ for some integer $q$. Then,

$$\frac{(a-b)\gcd(x,n)}{nx}=pb_1+qb_2$$

So, we conclude that $\dfrac{(a-b)\gcd(x,n)}{nx}$ is an integer, i.e, $\dfrac{(a-b)/x}{n/\gcd(x,n)}$ is an integer, so $\frac ax-\frac bx$ is divisible by $n/\gcd(x,n)$ which shows that $a/x\equiv b/x\pmod{n/\gcd(x,n)}$

More simply, write $\ \bar a = a/x,\ \bar b = b/x\$ and $y = \bar a - \bar b.\$ Then we get a $1$-line proof:

$$\,n\mid a\!-\!b=xy\iff n\mid xy,ny\color{#0a0}\iff n\mid (xy,ny)\!\color{#c00}{\overset{\rm D}{=}}\!(x,n)y \iff n/(x,n)\mid y$$

We employed the universal property $\color{#0a0}\iff$ along with the $\,\color{#c00}{\overset{\rm D}{=}}$ Distributive Law of the gcd.

Your proof looks correct. An easier way to do the sufficiency statement would be as follows:

We have $n|(a-b)$ and $x|(a-b)$, therefore if we let $L= lcm(n,x)$ then $L|(a-b)$. Now let $a-b = Lk$ for an integer $k$. Now divide through by $x$ and use the fact that $nx = gcd(n,x)lcm(n,x)$ to get $(a-b)/x = Lk/x = \frac{n}{gcd(n,x)} k$, which implies $a/x=b/x$ mod $\frac{n}{gcd(n,x)}$

Assume that $a \equiv b \pmod n$. Then there exist an integer $k$ such that $a-b = kn$. Now as $x$ divides the RHS we will have $\frac{x}{gcd(x,n)} \mid k$. So now we have:

$$\frac{a}{x} - \frac{b}{x} = \frac{k}{\frac{x}{gcd(x,n)}} \cdot \frac{n}{gcd(x,n)}$$

So as $\frac{k}{\frac{x}{gcd(x,n)}} \in \mathbb{Z}$ we have that $\frac{a}{x} \equiv \frac{b}{x} \pmod {\frac{n}{gcd(x,n)}}$

Now assume that $\frac{a}{x} \equiv \frac{b}{x} \pmod {\frac{n}{gcd(x,n)}}$. Then there exist integer $k$ such that $\frac{a}{x} - \frac{b}{x} = k \cdot \frac{n}{gcd(x,n)}$. Multiply by $x$ and you will get:

$$a - b = \left(k \cdot \frac{x}{gcd(x,n)}\right) \cdot n \implies a \equiv b \pmod n$$

Therefore as we proved both sides we get that:

$$a \equiv b \pmod n \iff \frac{a}{x} \equiv \frac{b}{x} \pmod{\frac{n}{gcd(x,n)}} \quad \text{, where: }x=gcd(a,b)$$