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This is an example to illustrate the following theorem:

Let $a,b$ be integers and let $m,n$ be positive integers. The system \begin{eqnarray*} x&\equiv & a\pmod{m}\\ x&\equiv & b\pmod{n}\\ \end{eqnarray*} has solution if and only if $(m,n)\mid (a-b).$ In this case, the solution can be uniquely expressed modulo $[m,n]$

Determine the solutions to the following linear system of equations \begin{eqnarray*} x&\equiv & 5\pmod{6}\\ x&\equiv & 8\pmod{15}\\ \end{eqnarray*}

The system will have a solution if $(15,6)\mid (5-8)$, wich is true, since $3\mid-3$. I have found that the solution of this system is $x\equiv 23\pmod{30}$, Which they argue as follows:

Since there is a solution, we express $-3$ as a linear combination of $6$ and $15$: $5-8= -3 = (-3)\cdot 6 + 1\cdot 15$, thus $x=8+15=23$.

In this step I am stuck, Is there another way to get to that 23?, if someone could explain in detail what was done in this step I would be very grateful.

Hopmaths
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1 Answers1

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The idea of the given step is as follows: First, subtract $8$ to get the following equations for $y=x-8$: $$y\equiv-3\pmod6\\ y\equiv0\pmod{15}$$

By the second equation, $y$ is of the form $15a, a\in\mathbb Z$. By the first one $y$ has the form $-3-6b$. Thus $-3-6b=y=15a$ means $-3=15a+6b$, which explains why we want to write $-3$ as a linear combination.

Tesla Daybreak
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