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Origin page 5. The overhead doesn't look like Linear Congruence Theorem or anything from Euclid's Algorithm. page 4 tries to delineate Casting Out the Modulus? Is this it?

So if $d|m,$ then I rewrite $\begin{align} ax & \equiv b \mod m \\dx + x & \equiv \\ \implies x & \equiv \end{align}$

Is this errorless? Did I blight anything?

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Dwayne E. Pouiller
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  • Finding an "appropriate factor" or "modular inverse" is, in general, solved by applying a version of Fermat's little theorem or the extended euclidean algorithm, so this is not different. Of course you can find the inverse also by trial or writing down the multiplication table if it is small enough. – Lutz Lehmann Jun 01 '14 at 22:24

2 Answers2

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The $3x \equiv 9 \mod 6 $ has a common factor of $3$. Dividing both sides by 3 - including the modulus - we get:

$$ \begin{array}{ccc} 3x & \equiv & 9 \mod 6 \\ x & \equiv & 3 \mod \mathbf{2} \end{array} $$

For the remaining two congruence, we multiply both sides by an appropriate number:

$$ \begin{array}{rcc} 2x & \equiv & 1 \mod 5 \\ 3(2x) & \equiv & 3 \mod 5 \\ x & \equiv & 3 \mod 5 \end{array} $$

In modular arithmetic we have inverses $3\times 2 = 6 \equiv 1 \mod 5$ so we can say that $2^{-1} = 3$.

If you are still not sure about $x \equiv 3 \mod 2$. Let's write the congruence as an actual equation:

$$ \begin{array}{ccc} 3x -9 & \equiv & 6k \\ x-3 & \equiv & 2k \end{array} $$

The modulus $6$ got reduced to a $2$ in the process.

cactus314
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2

I use the following strategy for solving the linear congruence

$$ax \equiv b \mod{m}\quad (\star)$$

Case 1: $\gcd(a, m) = 1$

Implies that the modular multiplicative inverse $(\color{green}{a^{-1}})$ exists, and we solve for $x$ as follows.

$\begin{eqnarray}&&ax &\equiv b&\mod{m}\\\iff&(\color{green}{a^{-1}})&ax &\equiv (\color{green}{a^{-1}})b&\mod{m}\\\iff&&x &\equiv \color{green}{a^{-1}}b&\mod{m}\tag{1}\end{eqnarray}$

Example: $$2x \equiv 1 \mod{5} \iff x \equiv 3 \mod{5}$$ where $\color{green}{2^{-1} = 3}$ was found using the Extended Euclidean algorithm.

Case 2: $\gcd(a, m) > 1$ and $\gcd(a, m) \nmid b$

$\text{No solution exists}\tag{2}$

$\begin{eqnarray}\text{since}&ax &\equiv b \mod{m}\\\iff &ax &= b + mt &(t \in\mathbb{Z})\\\iff &b&= ax - mt \end{eqnarray}$

so $b$ being a linear combination of $a$ and $m$, must be divisible by $\gcd(a, m)$.

Example: $$2x \equiv 3 \mod{4}$$ has no solution, since $\gcd(2, 4) = 2 \nmid 3$.

Case 3: $\gcd(a, m) > 1$ and $\gcd(a, m) \mid b$

Substitute $a = a_0d, b = b_0d, m = m_0d$ in $(\star)$ where $d = \gcd(a, m)$

$\begin{eqnarray}&ax &\equiv b &\mod{m}\\\iff&a_0dx &\equiv b_0d &\mod{m_0d}\\\iff &a_0dx &= b_0d + m_0dt &(t \in\mathbb{Z})\\\iff &a_0x &= b_0 + m_0t &(\text{divide by }d \neq 0) \\\iff&a_0x &\equiv b_0&\mod{m_0}\tag{3}\end{eqnarray}$

which finds itself amenable to solution by invoking case 1, since $\gcd(a_0, m_0) = 1$.

Examples:

  • $4x \equiv 2 \mod{10} \underset{(\text{case }3)}{\overset{\gcd(4, 10) = 2}{\iff}} 2x \equiv 1 \mod{5}\underset{(\text{case }1)}{\overset{\gcd(2, 5) = 1}{\iff}} x \equiv 3 \mod{5}$
  • $3x \equiv 9 \mod{6} \quad\underset{(\text{case }3)}{\overset{\gcd(3, 6) = 3}{\iff}} x \equiv 3 \mod{2}\quad\underset{(\text{case }1)}{\overset{\text{trivially}}{\iff}} x \equiv 3 \mod{2}$
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