RMO 2012 solution to Question No. 6

6. Find all positive integers n such that \(3^{2n} + 3 n^2 + 7 \) is a perfect square.

Solution:

We use the fact that between square of two consecutive numbers there exist no perfect square. That is between \(k^2 \) and \((k+1)^2 \) there is no square.

Note that \(3^{2n} = (9^n)^2 \) and  \((9^n + 1)^2 \) are two consecutive perfect square and \(3^{2n} + 3 n^2 + 7 \) is always a number between them for n > 2 (easily proved by induction).

Hence the only solution is n = 2.