Diophantine Equations are any equations where the values of the variables are required to be integers. See below for some examples.
General Solutions
For equations of the form $ax + by = c$, where $a, b, x, y \in \mathbb{Z}$ and $c \in \mathbb{N}$, we can use the following to find all solutions in $x, y$ (assuming there is a solution).
Step One: Find an initial answer for $x,y \in \mathbb{Z}$.
$(a)$ Solve $ax +by = c$ for $y$.
(1)$(b)$ Plug in integer values for $x$ until you get an initial integer value for $y$ in the equation $y=\frac{c-ax}{b}$. Call these initial values $x',y'$.
Step Two: Simplify $a,b \in \mathbb{Z}$ by dividing them by their GCD and call the simplified values $a',b'$.
$(a)$ Find GCD of $a,b$ and call it $d \in \mathbb{Z}$.
$(b)$ Divide $a,b$ by $d$ to get $a',b'$.
$a' = \frac{a}{d}$
$b' = \frac{b}{d}$
Step Three: Using the new values $x',y',a',b'$ you can find all integer values $x,y$ by plugging $x',y',a',b'$ into the following equations.
$(a)$ $x=(x'+b'n)$ when $y=(y'-a'n)$ where $n \in \mathbb{N}$
$(b)$ $x=(x'-b'n)$ when $y=(y'+a'n)$ where $n \in \mathbb{N}$
Example: $6x + 8y = 30$
Step One: $y=\frac{15-3x}{4}$
$x' = 1$
$y' = 3$
Step Two: $\gcd(6,8) = 2$
$a' = \frac{6}{2} = 3$
$b' = \frac{8}{2} = 4$
Step Three: Find all Solutions For $x,y \in \mathbb{Z}$
$x=(1+4n)$ when $y=(3-3n)$ where $n \in \mathbb{N}$
$x=(1-4n)$ when $y=(3+3n)$ where $n \in \mathbb{N}$