Can I solve for all integer solutions of this diophantine equation?

Question

I do not know much about this subject, but this problem is bothering me.

$$ x + 33y = 2399 $$

How can I find the possible integer values of x and y? I know there are two solutions, which I discovered by brute force. I just want to know if it's possible to find them without brute force.

Thanks.

edit: I tried to learn how to solve these types of equations, and I read they are solvable if gcd(a, b) can divide c. gcb(1, 33) is 1, and 1 divides 2399... Any insight?

Answer 1

One method of solution is as follows.

The equation $x + 33 y = 2399$ can be seen to be satisfied by: \begin{align} 2399 &= 23 + 33 \cdot 72 \\ &= (1 \cdot 33 + 23) + 33 \cdot 71 \\ &= (2\cdot 33 + 23) + 33 \cdot 70 \\ &= \cdots \end{align} This pattern yields a set of solutions $(x_{n}, y_{n})$ in the form \begin{align} (x_{n}, y_{n}) = (33\, n + 23, 72 - n) \hspace{10mm} 0 \leq n \leq 72. \end{align}

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A second method is as follows.

Let $x = u^{2} + 22$ and $y = 72 - v^{2}$ to obtain \begin{align} u^{2} - 33 \, v^{2} = 1. \end{align} This is a standard Pell equation and has the solution set \begin{align} u_{n} &= \frac{1}{2} \, \left[ (23 + 4 \sqrt{33})^{n} + (23 - 4 \sqrt{33})^{n} \right] \\ v_{n} &= \frac{1}{2 \, \sqrt{33}} \, \left[ (23 + 4 \sqrt{33})^{n} - (23 - 4 \sqrt{33})^{n} \right] \end{align} For reference see MathWorld Pell equation. Squaring these values one may obtain the desired set of $(x,y)$ solutions in the form $(x_{n}, y_{n}) = (u_{n}^{2} + 22, 72 - v_{n}^{2})$.

Answer 2

The first thing to notice about this equation is that for any integer value of $y$, there is an integer value of $x$ that solves the equation. If we set $y$ to some integer $n$, then the corresponding value of $x$ is $2399-33n.$ Hence the set of integer solutions consists of ordered pairs of the form $(x,y)=(2399-33n,n).$

Note that if we set $n$ to $72-t$ we get the equivalent parameterization, $(33t+23,72-t),$ recovering an earlier answer.