I don't know what approach they seek, but my approach (potentially going beyond year 9 or 10 Maths*) is as follows:

Rearranging the given equation to make

the subject, I get that

and sketching this gives a rectangular hyperbola with asymptotes at

and

.

View attachment 32730
From this graph, a few things are immediately clear:

- as increases, decreases
- we can only have
*both* and for and hence the minimum possible value for is 3.
- as increases, approaches the value 1.5, and so I can find an upper bound for by examining the case :

- So, if then (as decreases when increases), and since is bounded below by 1.5 for positive , cannot be an integer if . In other words, all solutions must satisfy .
- We already have one solution, and , for which

We could now find all solutions by trial-and-error by sequentially testing

. This gives us that

and that

, and thus

So, there are two solutions:

If the requirement that

and

be positive is removed, but they are still required to be integers, two other solutions appear:

but it does not add any extra possible values for

, which can only be -7 or 9.

*** A note on "potentially going beyond year 9 or 10 Maths"...** I originally defined a function

which has an oblique asymptote of

and then used the fact that

must be an integer to constrain the possible domain of

... once

is sufficiently close to the asymptote, any integer value of

will give a non-integer value of

. As I was typing this up, however, I realised that I could simplify the method by looking at just the above hyperbola, which makes the level of the Maths involved suited to junior high school so long as hyperbolas have been covered.