$\!\!\bmod n\!:\,$ a polynomial has a root $\!\iff\!$ it has a root in a complete residue system I'm reading some notes on discrete math and I can't follow this reasoning:

Let's see if there exist $x \in \mathbb{Z}$ such that $4x\equiv 3\pmod 6$. First let us observe that if $x\equiv x'\pmod 6$, $4x\equiv 4x'\pmod 6$ and then $x$ is a solution if and only if $x'$ is a solution. Therefore, just look for solutions in the set {$0,1,2,3,4,5$}.

I don't understand how any of this

$x\equiv x'\pmod 6$, $4x\equiv 4x'\pmod 6$ and then $x$ is a solution if and only if $x'$ is a solution.

is implying this

just look for solutions in the set {$0,1,2,3,4,5$}

I would appreciate any help.
 A: Note that $\gcd(4,6)=2$ so $4x$ is always even modulo $6$ thus cannot be $3$.
Now regarding the question, there are only $6$ elements in $\mathbb Z/6\mathbb Z$ which are $\{0,1,2,3,4,5\}$.
Any $x\in\mathbb Z$ can be reduced to one of this remainders modulo $6$:

*

*Euclidean division there exists $k$ and $0\le x'< 6$ such that $x=6k+x'$

Now since $4x=4(6k+x')=6(4k)+4x'\quad$ then $\quad 4x\equiv 4x'\pmod 6$
This means that you don't need to examine all integers $x$ but only the remainders $x'$ to solve the equation.
A: The key idea is quite simple, viz. $ $ an integer coefficient polynomial  $\,f(x)\,$ has a root $\!\bmod n\iff $ it has a root in a complete residue system $\!\bmod n,\,$ analogous to: $ $ a polynomial has a rational root $\iff$ it has a rational root in least terms. Let's examine this equivalence more closely.
Recall that the algebraic essence of a congruence is that it is not only an equivalence relation (generalized equality) but, furthermore, is compatible with the ambient algebraic operations, here (ring) addition and multiplication, i.e. it satisfies the following Congruence Sum and Product Rules
$$\begin{array}{}  a\equiv \bar a\\ b\equiv \bar b\end{array} \ \ \Rightarrow\ \ \begin{align} a+b\,&\equiv\, \bar a + \bar b\\ a \,*\, b\,&\equiv\, \bar  a\,*\, \bar b\end{align}\  \pmod{\!n}$$
By induction this extends to algebraic expressions composed of sums and products, e.g.  to polynomials $\,f(x)\,$ with integer coefficients, yielding the following Polynomial Congruence Rule
$$\begin{align} a\equiv \bar a\ \ \Rightarrow\ \ f(a)&\equiv f(\bar a)\ \pmod {\!n}\\[.3em]
\text{thus }\ f(a)\equiv 0\ \ \ &\!\!\!\!\iff f(\bar a)\equiv 0\end{align}$$
Thus to find a modular root of an integer  polynomial congruence it suffices to restrict attention to "normalized" roots $\,\bar a\,$ from any complete residue system, e.g. the standard  $\,\{0,1,2,\ldots n-1\}\,$ or $\,\{0,\pm1,\pm2,\ldots\} = \,$ least magnitude remainders, as in our above analogy: it suffices to search for rational roots in least terms,  i.e. "normalized" or "reduced" roots (e.g. as in the Rational Root Test).
In particular this is true for your polynomial $\,f(x) := 4x-3\pmod{\!6},\,$ i.e. it has a root $\,x\equiv a\pmod{\!6}\iff$ it has a root $\,\bar a\in \{0,1,\ldots 5\} = $ standard least nonnegative complete residue system $\bmod 6,\,$ exactly as claimed.
Remark for readers knowing a little algebra. The analogous fact that fraction equivalence is compatible with addition and multiplication is usually not even mentioned in elementary contexts but does require proof to be rigorous. This usually does not occur until a course in abstract algebra (when studying geneal fraction fields or localizations), but even then it is often "left to the reader".
