# Linear Diophantine Equation/Examples/5x + 6y = 1

Jump to navigation
Jump to search

## 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 $\text {2-13}$