How can I solve the folowing Diophantine equation with two unknowns?

Solution 1:

Hint $\ $ This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product, using the AC-method, viz.

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

So the problem reduces to checking which factors of $\,ad+bc\,$ have above form, a finite process.

Solution 2:

$\dfrac{274+y}{6y+1}=x$

As $x$is an integer, if integer $d$ divides both $274+y,6y+1;d$ must divide $6(274+y)-(6y+1)=1643$

For integer $x,6y+1$ must divide $1643=31\cdot53$ whose divisors are $\pm1\pm31,\pm53$

Solution 3:

This is @Bill Dubuque's answer in more detail.

I don't know what you mean by $\mathbb Z_0$.

$$ \forall x,y \in \mathbb Z_0, \quad 6xy + x - y = 274 $$

\begin{align} 6xy + x - y &= 274 \\ 36xy + 6x - 6y &= 1644 \\ 36xy + 6x - 6y - 1 &= 1643 \\ 6x(6y+1) -1(6y+1) &= 1643 \\ (6x-1)(6y+1) &= 1643 \end{align}

\begin{array}{|cc|cc|} \hline 6x-1 & 6y+1 & x &y \\ \hline 1 & 1643 & \ast & \ast\\ 31 & 53 & \ast & \ast\\ 53 & 31 & 9 & 5\\ 1643 & 1 & 274 & 0\\ \hline \end{array}