What's The Best Way To Factor This?

Relevant wiki: Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick

Rewrite the given equation as $xy - 3x - 2y = 0 \Longrightarrow (x - 2)(y - 3) = 6$ .

We then note that as $6$ has 8 integer divisors, namely $\pm 1, \pm 2, \pm 3$ and $\pm 6$ , we can express $6$ as a product of 2 integers in 8 ordered ways, i.e.,

• $x - 2 = 1, y - 3 = 6 \Longrightarrow (x,y) = (3,9)$ , or $x - 2 = 6, y - 3 = 1 \Longrightarrow (x,y) = (8,4)$ ,

• $x - 2 = 2, y - 3 = 3 \Longrightarrow (x,y) = (4,6)$ or $x - 2 = 3, y - 3 = 2 \Longrightarrow (x,y) = (5,5)$ ,

• $x - 2 = -1, y - 3 = -6 \Longrightarrow (x,y) = (1,-3)$ or $x - 2 = -6, y - 3 = -1 \Longrightarrow (x,y) = (-4,2)$ ,

• $x - 2 = -2, y - 3 = -3 \Longrightarrow ((x,y) = (0,0)$ or $x - 2 = -3, y - 3 = -2 \Longrightarrow (x,y) = (-1,1)$ .

Of these 8 cases, only 4 are such that both $x$ and $y$ are positive, namely $(3,9), (8,4), (4,6)$ and $(5,5)$ , so the desired sum is $(3 + 8 + 4 + 5) + (9 + 4 + 6 + 5) = 20 + 24 = \boxed{44}$ .

$3x+2y=xy$

$3x-xy+2y=0$

$(3-y)x=-2y$

$x=\frac{2y}{y-3}$ when $y$ is not equal to 3.

Let's look at the case $y=3$ :

$3x+2\times3=x\times3$

$6=0$ , which is false, so $y$ can't be equal to 3.

When $y=1$ or $y=2$ , the denominator $y-3$ in the equation $x=\frac{2y}{y-3}$ is negative, which means $x$ can't be a positive integer. Now, let's look at two inequalities when $y>3$ :

$x=\frac{2y}{y-3}>2$

$2y>2(y-3)$

$0>-6$ , which holds true with all $y>3$ .

Secondly,

$x=\frac{2y}{y-3}<3$

$2y<3y-9$

$y>9$ .

By looking at the two inequalities, $x$ is between 2 and 3 with all $y>9$ , meaning $x$ can't be an integer. This means there are no pairs of positive integer roots with $y>9$ . Now we are looking at $x$ -values with integer $y$ equal to or less than 9:

$y=4$ : $x=\frac{2\times4}{4-3}=8$

$y=5$ : $x=\frac{2\times5}{5-3}=5$

$y=6$ : $x=\frac{2\times6}{6-3}=4$

$y=7$ : $x=\frac{2\times7}{7-3}=\frac{7}{2}$

$y=8$ : $x=\frac{2\times8}{8-3}=\frac{16}{5}$

$y=9$ : $x=\frac{2\times9}{9-3}=3$

We see that there are four pairs of positive integer roots: $(4;8),(5;5),(6;4),(9;3)$ The asked sum is therefore $(4+5+6+9)+(8+5+4+3)=\boxed{44}$ .

For positive $x$ and $y$ , $3x + 2y = xy \Rightarrow \dfrac 3y + \dfrac 2x = 1 \Rightarrow x > 2, y > 3$ .
Also, $3x + 2y = xy \Rightarrow (x - 2)(y - 3) = 6$ .
Now $(x - 2)$ can be any positive factor of $6$ and so can $(y - 3)$ , since $\dfrac 6{x-2} = y-3$ .

Now the sum of the possible values for $x$ must be the sum of factors of $6$ plus $2$ for each $x$ , since we subtract $2$ .
Also the sum of the possible values for $y$ must be the sum of factors of $6$ plus $3$ for each $y$ , since we subtract $3$ .

Hence, the answer must be $2(1 + 2 + 3 + 6) + 4(3+2) = \boxed{44}$ .