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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.

Bill Dubuque
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xtreyreader
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  • Every integer is congruent modulo $6$ to one of $0$, $1$, $2$, $3$, $4$, or $5$. So if there is any solution, then there is a solution from among them and conversely. – Arturo Magidin Sep 28 '20 at 19:47

2 Answers2

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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.

zwim
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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, further, it is compatible with the ambient algebraic operations, here (ring) addition and multiplication, i.e. it satisfies these 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 sll algebraic expressions composed of sums and products, i.e. 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 (and count) modular roots of an integer polynomial congruence it suffices to restrict attention to the "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 congruent roots are counted only once since a complete residue system contains only one rep from each equivalence class, i.e. the elements of the system are incongruent, so we count at most one of $\,{-1}\equiv 3\,$ as a root of $\,x^2\equiv 1\pmod{\!4},\,$ just like we count at most one of $\,\frac{1}{2}\equiv \frac{5}{10}$ as a rational root.

Applying the above to your polynomial $\,f(x) := 4x-3\pmod{\!6},\,$ we infer: 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".

Bill Dubuque
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