Solving a Congruence - cannot understand a step in the solution New to congruences & Number Theory
Below is text from the book Joseph H. Silverman: A Friendly Introduction to Number Theory, 4th Edition, chapter 8, page 56.

To solve
$4x\equiv 3 \pmod{19}$
we will multiply both sides by $5$. This gives
$20x\equiv 15 \pmod{19}$ - Step 1
But $20\equiv 1\pmod{19}$, so $20x\equiv x\pmod{19}$ - Step 2
Thus the solution is
$x\equiv 15\pmod{19}$

I understand up to step 2, I am unable to understand how one arrives at the solution from Step 2.
How does
$20x\equiv x \pmod{19}$
lead to
$x\equiv 15 \pmod{19}$
Where did the $20$ on the LHS go? How did $x$ on the RHS get replaced by $15$?
 A: I think the issue here concerns the basic properties of congruence.
In many important ways, congruence behaves exactly like equality.  That is, it satisfies the three critical properties:
$1)$ Reflexive:  $a\equiv a \pmod n$.
$2)$ Symmetric:  $a\equiv b \pmod n\iff b\equiv a \pmod n$
$3)$ Transitive: $a\equiv b\pmod n$ and $b\equiv c\pmod n$ imply $a\equiv c \pmod n$.
Each of these follow easily from the core definition of congruence.
Those three properties, together, make congruence an Equivalence Relation.  That's an important notion on its own..,in many ways, you can work with Equivalence Relations the same way you work with Equality.  That's what is going on in the given calculation.
In this case you have $$20x\equiv x\pmod {19}\quad \&\quad 20x\equiv 15\pmod {19}$$ so combining the Symmetric Property and the Transitive Property get us $x\equiv {15}\pmod {19}$.
As usual, though, the important thing is the general principle.  Those three properties are why congruences are so useful and important...make sure you understand why they hold.
A: I will stress that $\gcd(5,19)=1$. Since $5$ is coprime to the modulus, multiplying by $5$ does not change the solutions so these two congruences are equivalent1
$$4x\equiv3\pmod{19} \Longleftrightarrow 20x\equiv15\pmod{19}$$
Now since $x\equiv20x\pmod{19}$, the latter is equivalent to $x\equiv15\pmod{19}$.

Since the comments here (and to the other answers) clarified that this is the main problem, let me spell the last equivalence in detail. (I will be freely using both symmetry and transitivity.)

*

*$x\equiv20x\pmod{19}$ and $20x\equiv15\pmod{19}$ implies $x\equiv15\pmod{19}$

*$20x\equiv x\pmod{19}$ $x\equiv15\pmod{19}$ implies $20x\equiv15\pmod{19}$

*So we have both $$20x\equiv15\pmod{19} \Longrightarrow x\equiv15\pmod{19}$$ and $$x\equiv15\pmod{19} \Longrightarrow 20x\equiv15\pmod{19}$$
which gives us the equivalence $x\equiv15\pmod{19} \Longleftrightarrow 20x\equiv15\pmod{19}$.


1See, for example:

*

*Proving the equivalence between two congruences.

*Prove that multiplication by an integer $a$ that is relatively prime to $n$ defines a bijection from $\mathbb{Z}_n-\{0\}$ to itself

As a side note, I will mention that there exist chatrooms such as Number Theory and Basic Mathematics. And there is also the main chatroom for this site.
See also: List of chatrooms. (I am mentioning this mainly since I saw that you had several exchanges in comments. If there are too many comments, that might be a sign that discussion in chat might be more suitable.)
A: Well, $20\equiv 1 \mod 19$ and so $20\cdot x\equiv 1\cdot x\mod 19$.
The rest is how you explained it: Multiplying $4x\equiv 3\mod 19$ by $5$ on both sides gives $20x\equiv 15\mod 19$, i.e., $x\equiv 15\mod 19$.
A: From here
$$20x\equiv 15 \mod19$$
we have that
$$20x=19x+x \implies 20x\equiv x \mod19$$
therefore
$$20x\equiv x\equiv 15 \mod19$$
Indeed by definition
$$a\equiv b \mod n \iff a-b=kn$$
therefore $20x\equiv x \mod 19 $ since $20x-x=19x$.
A: ou can divide sides of relation resulted in step 1 be sides of relation resulted in step 2:
$\frac{20x}{20x}  ≡ \frac {15} x \mod (19)$
⇒ $1 ≡ \frac {15} x \mod (19)$
⇒ $x ≡ 15 \mod (19)$
