Let the total cost of the sweets be \(c\) pence for \(n\) bags at \(p\) pence, \(n + 2\) bags at \(p – 9\) pence and \(m\) bags at \(p – 12\) pence. Hence:\[c=n p=(n + 2)(p – 9)=m(p-12)\]Eliminating \(c\) gives:\[\begin{eqnarray}p\;=\;9(n+2)/2\; =\;12m/(m-n)\end{eqnarray}\]Now eliminating \(p\) and arranging the result gives:\[(3m – 10)^2-(6n-3m+6)^2=64\]Denoting the squared terms as \(v\) and \(w\) we have \[v^2 – w^2 = (v – w)(v + w) = 64\] With\(f\) as any divisor of \(64\), we can now equate factors on either side of this equation to obtain: \[\begin{eqnarray}v-w\;&=&f\\v+w\;&=&64/f\end{eqnarray}\] Solving for \(v\) and \(w\) now gives:\[\begin{eqnarray}2v\;&=&64/f+f\;&=&6m-20\\2w\;&=&64/f-f\;&=&12n-6m+12\end{eqnarray}\] and hence:\[\begin{eqnarray}n\;&=&(32/f+2)/3\\m\;&=&n+(f+16)/6\end{eqnarray}\]There are only a few divisors of 64 to try and \(f=2\) produces a solution.

It is also easy to derive an alternative formulation which gives: \[\begin{eqnarray} p\;&=&\;9k \\ n\;&=&\;2(k-1) \\ c\;&=&\;18k(k-1) \\
m\,&=&\;6k(k-1)/(3k-4)\end{eqnarray}\]for integer \(k\), which gives the solution for \(k=4\)