Overview

A Diophantine equation is an algebraic equation where we are strictly looking for integer solutions. The most fundamental type is the Linear Diophantine Equation, which takes the form $ax + by = c$, where $a, b, c$ are given integers, and $x, y$ are unknown integers.

Geometrically, the equation $ax + by = c$ represents a line in the Cartesian plane. Solving the Diophantine equation means finding the exact points on this line where both the x-coordinate and y-coordinate are integers (often called lattice points).

Key Ideas

• Existence of Solutions (Bézout's Identity): The equation $ax + by = c$ has integer solutions if and only if the greatest common divisor $\gcd(a, b)$ divides $c$. If $\gcd(a, b) \nmid c$, no integer solutions exist.
• Finding a Base Solution: If the numbers are small, guess and check. If they are large, use the Extended Euclidean Algorithm to work backward and find a base solution $(x_0, y_0)$.
• Generating the Infinite Solution Set: Once you have one valid point $(x_0, y_0)$, you can find all other integer points on the line using the slope. The general solution is: $x = x_0 + \frac{b}{\gcd(a, b)} \cdot k$ $y = y_0 - \frac{a}{\gcd(a, b)} \cdot k$ where $k$ is any integer. Notice how as $x$ steps forward by the reduced $b$ coefficient, $y$ must step backward by the reduced $a$ coefficient to keep the sum balanced.

Worked Examples

Example 1: Finding the General Solution

Find all integer solutions to $12x + 15y = 39$.

1. Check for existence: First, compute $\gcd(12, 15) = 3$. We check if $3 \mid 39$. Since $39 / 3 = 13$, integer solutions exist.

2. Simplify the equation: Divide the entire equation by the GCD to make the coefficients coprime: $4x + 5y = 13$

3. Find a base solution: By inspection, $4(2) + 5(1) = 8 + 5 = 13$. So, $(x_0, y_0) = (2, 1)$ is a valid base solution.

4. Write the general solution: Using the simplified coefficients $a'=4$ and $b'=5$, we apply the general formula. As $x$ increases by $5$, $y$ must decrease by $4$: $x = 2 + 5k$ $y = 1 - 4k$ where $k \in \mathbb{Z}$.

Example 2: Recognizing When No Solutions Exist

Does the equation $14x + 21y = 50$ have any integer solutions?

1. Check for existence: Compute $\gcd(14, 21) = 7$. We then check if $7 \mid 50$. Since $50$ is not a multiple of $7$, Bézout's Identity tells us that it is impossible to form the sum of $50$ using integer multiples of $14$ and $21$. Conclusion: There are no integer solutions.

Example 3: Word Problems and Real-World Constraints

A farmer buys chickens for $3$ dollars each and pigs for $5$ dollars each. If he spends exactly $34$ dollars, what are all the possible combinations of chickens and pigs he could have bought?

1. Set up the equation: Let $c$ be the number of chickens and $p$ be the number of pigs. $3c + 5p = 34$ Because he is buying physical animals, we have a real-world constraint: $c \ge 0$ and $p \ge 0$.

2. Isolate one variable: It is usually easiest to isolate the variable with the smaller coefficient. $3c = 34 - 5p \implies c = \frac{34 - 5p}{3}$

3. Test valid inputs: Since $c$ must be an integer, $34 - 5p$ must be a multiple of $3$. We test non-negative integers for $p$:

• If $p = 0$: $c = 34 / 3$ (Not an integer)
• If $p = 1$: $c = (34 - 5) / 3 = 29 / 3$ (Not an integer)
• If $p = 2$: $c = (34 - 10) / 3 = 24 / 3 = 8$. Valid solution: $(8, 2)$
• If $p = 3$: $c = (34 - 15) / 3 = 19 / 3$ (Not an integer)
• If $p = 4$: $c = (34 - 20) / 3 = 14 / 3$ (Not an integer)
• If $p = 5$: $c = (34 - 25) / 3 = 9 / 3 = 3$. Valid solution: $(3, 5)$
• If $p = 6$: $c = (34 - 30) / 3 = 4 / 3$ (Not an integer)
• If $p = 7$: $c = (34 - 35) / 3 = -1 / 3$ (We stop here because $c$ is now negative).

Conclusion: The farmer could have bought 8 chickens and 2 pigs, or 3 chickens and 5 pigs.

Common Pitfalls

• Forgetting to divide by the GCD: When writing the general solution, students often use the original coefficients (e.g., $+15k$ and $-12k$). You must divide $a$ and $b$ by $\gcd(a,b)$ to ensure you don't skip over valid integer points between your steps.
• Sign errors in the general solution: Remember that one variable must increase while the other decreases to maintain the constant sum $c$.
• Ignoring constraints: In word problems, variables often represent physical quantities that cannot be negative. Always bounds-check your general solution against $x \ge 0$ and $y \ge 0$ if the context demands it.

Practice Problems