Why does this equality hold when solving linear congruence?

Question

I am quite confused about specific equality that is used when solving linear congruences. In the example provided, it is the jump from the second to last equation to the final equation. I cannot seem to find any theorem or even reasoning as to why this works. I am trying to use the method of solving for the inverse, then using this inverse to find the answer.

For example, to solve

$3x \equiv 4$ $($$\mod 7$$)$

first, we can find the inverse of $3$ modulus $7$. This inverse is equal to $-2$.

Therefore, we can multiply both sides by $-2$, getting $(-2) \cdot3x \equiv (-2)\cdot4$ $($$\mod 7$$)$, or

$-6x \equiv -8$ $($$\mod 7$$)$

This is where I am confused. I don't understand why this allows us to say that since

$-6x \equiv -8$ $($$\mod 7$$)$,

$x \equiv -8$ $($$\mod 7$$)$

I have been looking through every proof I could find and I cannot figure why this second to last step implies the final step.

You said yourself inverse of $3$ is $-2$ so since $aa^{-1}\equiv 1$ we have $-6\equiv 3(-2)\equiv 1\pmod 7$. — zwim, Apr 24 '20 at 20:24
@zwim so with congruences, are we allowed to have a bunch of connected congruences like that? For example, -6≡1 (mod 7) ⇒ -6x ≡ x (mod 7) (by multiplying both sides by x), therefore -6x ≡ x ≡ -8 (mod 7) — Franklin V, Apr 24 '20 at 20:28
$\color{#c00}{-6\equiv 1},\Rightarrow,\color{#c00}{-6}x\equiv \color{#c00}1x\equiv x,$ by the Congruence Product Rule in the linked dupe. — Bill Dubuque, Apr 24 '20 at 20:52

score 2 · Accepted Answer · answered Apr 24 '20 at 20:30

Well if $a\equiv \bar a\pmod n$ then there exist $k$ such that $a=nk+\bar a$

If you multiply by $x$ then $ax=n(kx)+\bar ax=nK+\bar ax\quad$ for an integer $K=kx$.

Meaning $\quad ax\equiv \bar ax\pmod n$

So $-6\equiv 1\pmod 7\implies -6x\equiv 1x\pmod 7$.

Note: the reciprocal (i.e. division by $x$) is only true if $\gcd(x,n)=1$.

Why does this equality hold when solving linear congruence?

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