Determining number of integer solutions of Diophantine equation

Question

Question:

How many integer solutions (consisting of two integers x and y) does the equation below have? Note: You are not required to find any solutions if there are any.

$ 294x + 266y = 28$

My attempt:

I am not sure what the question means by "how many". We know that gcd$(294,266)$ divides $28$ so there are infinitely many solutions to this Diophantine equation.

Answer 1

Yes, ther are infinite solutions.

Given $294x + 266y = 28$, we can see that $\gcd(294, 266) = 14$ divides the answer, $28$, so solutions are feasible.

Dividing the equation through by $14$, we get $21x+19y=2$. Clearly $(x,y) = (1,-1)$ is a solution and for any solution $(x,y)=(k,\ell)$ we will have $(k+19, \ell-21)$ and $(k-19, \ell+21)$ will also be solutions. Hence there are an infinite number of solutions.

In general after the division we will have coprime coefficients so using Bézout's identity we can adjust the values of $x$ and $y$ to give any integer answer and then generate the infinite set similarly to above.

Answer 2

By a convoluted process we can find $$294x + 266y = 28\quad\implies x = -19 n - 18 \quad\land\quad y = 21 n + 20 \quad\land\quad n\in\mathbb{Z}$$ This can be verified in a spreadsheet showing the samples below

\begin{align*} &(\space n\quad x\quad y\quad294x + 266y - 28)\\ &(-4,58,-64,0)\\ &(-3,39,-43,0)\\ &(-2,20,-22,0)\\ &(-1,1,-1,0)\\ &(0,-18,20,0)\\ &(1,-37,41,0)\\ &(2,-56,62,0)\\ &(3,-75,83,0)\\ &(4,-94,104,0)\\ \end{align*} Note that the last element is alway zero meaning all $(x,y)$ shown are valid. Since these are all derived from the set of integers $(\mathbb{Z})$, there are infinite solutions to the equation.

Answer 3

$294x+266y=28$

Dividing by $14$ we get an equation with the same solutions

$21x+19y=2$

The equality $A=B$ implies that $A\equiv B\pmod m$ for any $m \in Z$

$21x+19y\equiv 2 (mod 19)$

$2x\equiv 2 (mod 19)$

$gcd(2;19)=1$

So we can divide by 2

$x\equiv 1 (mod 19)$

$x=19k+1$

We can substitute this into our equation

$21(19k+1)+19y=2$

$21\times19k+21+19y=2$

$19y=2-21\times19k-21$

$19y=-21\times19k-19$

$y=-21k-1$

$(19k+1; -21k-1)$ solves the equation for every $k\in Z$, therefore the equation has infinitly many solutions