MHB Solve Integer Equation: 3x+5y=2xy-1

[kaliprasad]

Solve in integers $3x + 5y = 2xy - 1$

 

[Opalg]

[sp]
Multiply by $2$ and rearrange: $4xy - 6x - 10y = 2.$

Factorise: $(2x-5)(2y-3) = 17.$

Since $17$ is prime, there are just four possible cases:

1) $\quad 2x-5 = 17$, $2y-3 = 1$, giving $(x,y) = (11,2).$

2) $\quad 2x-5 = 1$, $2y-3 = 17$, giving $(x,y) = (3,10).$

3) $\quad 2x-5 = -17$, $2y-3 = -1$, giving $(x,y) = (-6,1).$

4) $\quad 2x-5 = -1$, $2y-3 = -17$, giving $(x,y) = (2,-7).$
[/sp]

 

[kaliprasad]

[sp]
Multiply by $2$ and rearrange: $4xy - 6x - 10y = 2.$

Factorise: $(2x-5)(2y-3) = 17.$

Since $17$ is prime, there are just four possible cases:

1) $\quad 2x-5 = 17$, $2y-3 = 1$, giving $(x,y) = (11,2).$

2) $\quad 2x-5 = 1$, $2y-3 = 17$, giving $(x,y) = (3,10).$

3) $\quad 2x-5 = -17$, $2y-3 = -1$, giving $(x,y) = (-6,1).$

4) $\quad 2x-5 = -1$, $2y-3 = -17$, giving $(x,y) = (2,-7).$
[/sp]

My starting point is different and it becomes same as above

Spoiler

We have $2xy - 3x - 5y - 1 = 2( x - \frac{5}{2}) ( y - \frac{3}{2}) - 1 - \frac{15}{2} = 0$
or $ 2 ( 2x - 5) (2y - 3) = 2 * 17$
or $( 2x - 5) (2y-3) = 17$
Note that LHS is odd and so is RHS.
giving 4 solutions
$2x- 5 = -1, 2y -3= -17$ or $x = 2, y = - 7$
$2x -5 = -17, 2y -3= -1$ or $x = -6, y = 1$
$2x- 5 = 1, 2y -3= 17$ or $x = 3, y = 10$
$2x -5 = 17, 2y -3= -1$ or $x = 11, y = 2$