Linear Diophantine Equation/Examples/5x + 6y = 1
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Example of Linear Diophantine Equation
The linear diophantine equation:
- $5 x + 6 y = 1$
has the general solution:
- $x = -1 + 6 t, y = 1 - 5 t$
Proof
We have that:
- $\gcd \set {5, 6} = 1$
which is (trivially) a divisor of $1$.
So, from Solution of Linear Diophantine Equation, a solution exists.
First we find a single solution to $5 x + 6 y = 1$:
- $1 = 1 \times 6 - 1 \times 5$
So $y_0 = 1, x_0 = -1$ is a solution.
From Solution of Linear Diophantine Equation, the general solution is then:
- $\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
\(\ds x_0\) | \(=\) | \(\ds -1\) | ||||||||||||
\(\ds y_0\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds a\) | \(=\) | \(\ds 5\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds 6\) | ||||||||||||
\(\ds d\) | \(=\) | \(\ds 1\) |
giving:
\(\ds x\) | \(=\) | \(\ds -1 + 6 t\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds 1 - 5 t\) |
$\blacksquare$
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-3}$ The Linear Diophantine Equation: Example $\text {2-13}$