Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 5 - Objective Problem 5

The only possible values of $n$ are $n=4,7.$ For $n=4$ $(2,2)$ solves the equation, and for $n=7$ $(1,2)$ solves the equation.

So let's show that there are no other solutions. First note that for $n \le 3$ we can rewrite the equation as $a^2 + ab(1-n)+b^2 = -n,$ and for $n \le 3$ the left side is nonnegative but the right side is negative.

From now on, then, we can assume $n > 4.$ So now suppose the equation has positive integer solutions. Let $(a,b)$ be the minimal positive integer solution with $a \le b.$ If $a=b,$ then we get $3a^2 = n(a^2-1),$ or $a^2 = \frac{n}{n-3},$ but $1 < \frac{n}{n-3} < 4$ if $n > 4$ so this is impossible. So we can assume that $a < b.$ Now we use some Vieta jumping ideas: using the quadratic formula gives $b = \frac{a(n-1) \pm \sqrt{a^2(n-1)^2 - 4(a^2+n)}}2,$ and minimality implies that this particular $b$ is chosen using the negative sign. (It's the smaller of the two possible solutions, both of which must be positive integers since one of them is a positive integer by assumption, they sum to the positive integer $a(n-1),$ and their product is the positive integer $a^2+n.$ )

So rewrite: $\begin{aligned} a &< b \\ a &< \frac{a(n-1) - \sqrt{a^2(n-1)^2 - 4(a^2+n)}}2 \\ 2a &< a(n-1) - \sqrt{a^2(n-1)^2 - 4(a^2+n)} \\ a(n-3) &> \sqrt{a^2(n-1)^2 - 4(a^2+n)} \\ a^2(n-3)^2 &> a^2(n-1)^2 - 4(a^2+n) \\ 4(a^2+n) &> a^2(n-1)^2 - a^2(n-3)^2 \\ 4(a^2+n) &> a^2(4n-8) \\ 4n &> a^2(4n-12) \\ \frac{n}{n-3} &> a^2 \end{aligned}$ But again, for $n > 4,$ $\frac{n}{n-3}$ is strictly between $1$ and $4,$ so the only positive integer solution to this inequality is $a=1.$

In this case, we get $1+b+b^2 = n(b-1),$ and mod $b-1$ this simplifies to $3 \equiv 0,$ so $(b-1)|3,$ so $b=2,4.$ In both cases this gives $n=7.$ So we are done. There are exactly $\fbox{2}$ admissible values of $n.$