Can We Deal With Them Separately?

Relevant wiki: Quadratic Diophantine Equations - Solve by Factoring

$\dfrac{5(y-3)}{2}+\dfrac{18}{y+4}=\dfrac{5y^2+5y-24}{2(y+4)}$

$\because 2(y+4)$ is even, $5y^2+5y-24$ is even $\Rightarrow 5y^2+5y$ is even $\Rightarrow y(y+1)$ is even.

$\therefore y$ can be even or odd.

When $y$ is odd, $\frac{5}{2}(y-3)$ is an integer, hence $\frac{18}{y+4}$ is also an integer, so we consider integer factors of 18.

$\because y+4$ is odd, the solutions include all odd factors of 18, that gives us $\boxed{6}$ solutions.

When $y$ is even, $\frac{5}{2}(y-3)=\frac{k}{2}$ , where $k$ is an odd integer

$\therefore \frac{18}{y+4}=\frac{h}{2}$ , for some odd integer $h$ , so that the sum would be an integer.

Let $y+4=4r\Rightarrow\frac{18}{y+4}=\frac{18}{4r}=\frac{1}{2}\cdot\frac{9}{r}$

$\frac{9}{r}$ has to be an odd integer. $\therefore r=\pm1,\pm3,\pm9$ , which correspondingly gives us $\boxed{6}$ integer solutions of $y$ .

Therefore, we have $6+6=\boxed{12}$ integer solutions of $y$ .

Most people would think that we must have $y + 4 \mid 18$ , and thus $y + 4 = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$ . Then, checking that $\frac{ 5(y-3) } { 2}$ is an integer means that $y$ is odd, and thus we have $y + 4 = \pm 1, \pm 3, \pm 9$ which gives us 6 solutions.

However, they made the assumption that both of these terms must individually be integers, which need not be true. For example, with $y = 0$ , we get $\frac{ - 15 } { 2} + \frac{18}{4} = -3$ . We're missing out solutions where we have fractions summing up to integers, and these account for 6 more solutions making a total of 12.

How can we properly answer this problem?

---

Combining the fractions, we want $N= \frac{ 5y^2 + 5y -24}{ 2 (y+4) }$ to be an integer. If the denominator was only a monic linear term, we know how to deal with it using the remainder factor theorem . However, the leading coefficient of 2 introduces some complications for us.

Let's do some wishful thinking and get rid of it by multiplying by 2. IE $N= \frac { 5y^2 + 5y -24}{ 2 (y+4) }$ is an integer $\Rightarrow 2N= \frac { 5y^2 + 5y -24}{ (y+4) }$ is an (even) integer. (Note the single direction implication, where we get double implication if the integer is even). In this case, we know that we simply need $\frac{ 36}{y+4}$ to be an integer, or that $y+ 4 = \pm 1, 2, 3, 4, 6, 9, 12, 18, 36$ .
Now, with these possible values of $y$ , we test and see which ones of them give us an even $2N$ .

Note: Comparing with the other solutions, I do have a lot more values that need to be checked / rejected. However, as I do not have multiple cases, the initial thought analysis doesn't have to be reapplied to various cases/sub-cases.

My solution is merely streamlining some of the others.

If $y$ is odd, then $y+4$ must be an odd divisor of $18$ , so that $y+4=\pm1, \pm3,\pm 9$ .
If $y=2x$ is even, then $\frac{18}{y+4}=\frac{18}{2x+4}=\frac{9}{x+2}=\frac{k}{2}$ for an odd $k$ , meaning that $x+2$ is an even divisor of 18, so $x+2=\pm 2, \pm 6, \pm 18$ and $y+4=\pm 4, \pm 12, \pm 36$ . We have $\boxed{12}$ solutions in all.

For simplicity, let $y+4=x$ , so $\dfrac{5(y-3)}{2}+\dfrac{18}{y+4}$ becomes $\dfrac{5(x-7)}{2}+\dfrac{18}{x}$ . Also, let $L= \dfrac{5(x-7)}{2}$ and $R=\dfrac{18}{x}$ . Since the numerator of $L$ is an integer, and the denominator 2, $L$ will either end with a .5 or .0 after the decimal point. For the sum of $L$ and $R$ to be an integer, the $R$ must therefore end in either a .5 or a .0 also.

Then, it must be true that $\dfrac{18}{x}=\dfrac{I}{2}$ for some integer $I$ . It follows that $x$ is an integer factor of 36. So all satisfying values of $x$ are among $\pm1, \pm2, \pm3, \pm4, \pm6, \pm9, \pm12, \pm18, \pm36$ .

For $L$ to have no remainder, $5(x-7) \vert 2 \implies x$ is odd. Only a few values in the list are odd. Each one is a factor of 18, so $R$ also has no remainder. Therefore, $x=\pm1, \pm3, \pm9$ all yield integers.

If $x$ is even, $L$ has .5 occur after the decimal, so $R$ must as well. Then $x$ is not a factor of 18. Therefore, values $x=\pm4, \pm12, \pm36$ all yield integers.

There are therefore 12 distinct values of $x$ . Since $x-4=y$ , there are also 12 distinct values of $y$ .

PS: Rather than talking about numbers as ending with .5s or .0s, there is a more formal was uses $a-\left \lfloor{a}\right \rfloor$ to show mathematically everything I talk about with .5s combining to make integers etc, however, I believe it would have convoluted my solution, especially when my goal is simply to share my course of action, as I believe it can be easily understood exactly what I mean when I talk about .5s and .0s.