Find the general solution of the following Diophantine equations:
(a) $2x + 3y= 5$
(b) $3x + 6y= 9$
(c) $15x − 20y= 10$
For this exercise we make use of Proposition (1.17).
Let $\gcd (a, b) = g$. The Diophantine equation $ax + by= c$ has integer solutions if and only if $g \mid c$.
We also make use of Proposition (1.18).
Let $\gcd (a, b) = g$. If $g \mid c$ and $x_0, y_0$ are particular solutions of the equation $ax + by= c$, then all the other solutions of this equation are given by $$x = x_0 + (\frac{b}{g} )t \quad \text{and} \quad y= y_0 − (\frac{a}{g} )t$$ where $t$ is any integer.
(a) Here $g = \gcd(2,3)=1$ and so $g \mid 5$. Therefore, the equation has integer solutions, by Proposition 1.17.
By inspection $x_0=1$ and $y_0=1$ is a solution. The general solution, by Proposition 1.18, is
$$ x = 1 + 3t \quad y = 1 - 2t$$
for any integer $t$.
(b) Here $g = \gcd(3,6) = 3$ and so $g \mid 9$. Therefore, the equation has integer solutions, by Proposition 1.17.
By inspection $x_0 = 1$ and $y_0 = 1$ is a solution. The general solution, by Proposition 1.19, is
$$ x = 1 + 2t \quad y = 1 - t$$
for any integer $t$.
(c) Here $g = \gcd(15,20) = 5$ and so $g \mid 10$. Therefore, the equation has integer solutions, by Proposition 1.17.
By inspection $x_0 = 10$ and $y_0 = 7$ is a solution. The general solution, by Proposition 1.19, is
$$ x = 10 -4t \quad y = 7 - 3t$$
for any integer $t$.
Note: The author's solutions look different. The scaling of $t$ is the same, but the constants are different. We can shift the constants from $x$ and $y$ scaling according to the given equation to solve.