What exactly does "Solving a congruence" mean? I am new to Congruences.
I am reading through the chapters on Congruences in Joseph Silverman's Number Theory book & I am unable to figure out what exactly "solving a Congruence" means.
When you solve normal algebra equations with variables, the end result is usually getting a value for the variable(s).
What should be the end result for solving a congruence?
For e.g. if I solve
$4x\equiv 3(\mod19)$
What should I end up with?
Is it something like
$x\equiv \text{constant}_1(\mod \text{constant}_2)$?
Or is it something else? Is there a general description of what the solution should look like?
 A: I agree with Dietrich Burde's answer, but would like to express it differently, to expand the OP's intuition.
Let $R$ denote the set $\{0,1,\cdots,18\}.$
That is, $R$ is the complete set of residues mod 19.
This means that if $a$ is any positive integer, a non-negative integer $P$ can be found so that 
$a = [(19)\times P] + r,$ where $r \in R.$ 
In such a circumstance, $a$ is construed to be equivalent to $r$ mod 19.
Your original problem may be interpreted as
find all positive integers $x$ such that ($4 \times x$) is equivalent to 3, mod 19.
In actual fact, instead of checking every positive integer $x$ to see if it satisfies the modular equation, all you need to do is check every element in $R$.
This is because if $(4 \times r) \equiv 3 \pmod{19},$ and 
$s \equiv r \pmod{19}$ then 
$(4 \times s) \equiv 3 \pmod{19}.$
Given that, Dietrich Burde's answer employed the following general result from number theory:
Since 4 is co-prime to 19, there is exactly one element $r$ in $R$ 
such that $4 \times r \equiv 1 \pmod{19}.$
Once this value for $r$ is found, an immediate result is that
$4 \times (r \times 3) = (4 \times r) \times 3 \equiv (1 \times 3) \equiv 3 \pmod{19}.$
It just so happens that $4 \times 5 \equiv 1 \pmod{19}.$
Thus, $4 \times (5 \times 3) = (4 \times 5) \times 3 \equiv (1 \times 3) \equiv 3 \pmod{19}.$
Another general result from number theory is that once the satisfying residue of $5 \times 3 = 15$ is found 
[i.e. $4 \times 15 \equiv 3 \pmod{19}],$
this solution will be unique with respect to $R.$
That means that for any value $r$ in $R$ except $r = 15,$ 
$4 \times r$ will not be equivalent to 3 mod 19.
A: You should end up with a unique solution of $4x=3$ in the field $\Bbb F_{19}$, or formulated differently in an $x$, up to multiples of $19$ satisfying the congruence $x\equiv 15\bmod 19$.
Since $4$ is invertible in $\Bbb F_{19}$ with inverse $5$ (since $4\cdot 5=20=1$), we obtain
$$
x=4^{-1}\cdot 3=5\cdot 3=15.
$$
A: COMMENT.-The equation $ax\equiv b\pmod n$ means that $a$ and $b$ are elements of the ring $\mathbb Z/n\mathbb Z$ of classes of integers modulo $n$ and it is required to find the value (none, one or more) of $x$ that satisfy the equation $ax=b$ in this ring. In your example
$4x=3$ the class of $4$ modulo $19$ is invertible in the ring (a field in this case because $19$ is a prime) and its inverse is $5$ because $4\cdot5=20\equiv 1\pmod{19}$ so you can say that $x=5\cdot3=15$ in the $\mathbb Z/n\mathbb Z$ which is the same that $x$ is any integer of the form $15+19k$ for arbitrary integer $k$.
