NZMO Round One

Solution:
Let $f(n) = 4n^{4} - 4n^{3} + 12n^{2} + 20 = 4(n^{4} - n^{3} + 3n^{2} + 5)$ and note that $(n^{4} - n^{3} + 3n^{2} + 5)$ is a perfect square if and only if $f(n)$ is. First note that:
$$(2n^{2} - n + 5)^{2} - f(n) = 9n^{2} - 10n + 5 = 4n^{2} + 5(n - 1)^{2} > 0.$$
Also note that
$$f(n) - (2n^{2} - n + 2)^{2} = 3n^{2} + 4n + 16 = 2n^{2} + (n + 2)^{2} + 12 > 0.$$
Therefore $(2n^{2} - n + 2)^{2}< f(n)< (2n^{2} - n + 5)^{2}$, so the only way $f(n)$ could be a perfect square is if it is $(2n^{2} - n + 3)^{2}$ or $(2n^{2} - n + 4)^{2}$. Solving $f(n) = (2n^{2} - n + 3)^{2}$ gives us the quadratic $n^{2} - 6n - 11 = 0$ which has no integer solutions. Solving $f(n) = (2n^{2} - n + 4)^{2}$ gives us $5n^{2} - 8n - 4 = (5n + 2)(n - 2) = 0$, which has only one integer solution $n = 2$. Checking
$$(2)^{4} - (2)^{3} + 3(2)^{2} + 5 = 25$$
which is a perfect square. Therefore the only solution is $n = 2$. $\square$