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# 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.

## By Ashani Dasgupta

Founder Director at Cheenta
Pursuing Ph.D. in Mathematics from University of Wisconsin Milwaukee
Research Interest - Geometric Topology

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