RMO 2012 Problem

Solution outline:

$(3^n)^2<3^{2n}+3n^2+7<(3^n+1)^2$

for $n>2$ .

So the given expression can't be a perfect square for any $n>2$ . Checking $n=1, 2$ , we see that only $n=2$ works.

So our answer is $\boxed{2}$ .

This is longer solution, but good enough.

We check two cases about the property of $n$ .

(a) If n is odd, ie. $n = 2k+1$ for any non-negative integer $k$ , then $n^2$ is of the form $4m+1$ , for some non-negative integer $m$ . Thus the expression turns to

$3^{4k + 2}+3(4m + 1)+7=9 \cdot 81^{k} + 12m + 10$

Now, $9 \cdot81^{k}$ is congruent to $1\mod 4$ , giving $9 \cdot 81^{k} + 12m + 10$ congruent to $3\mod 4$ . But a perfect square can't be congruent to $3\mod 4$ . Thus, $n$ cannot be an odd positive integer.

(b) If n is even, ie. $n = 2k$ for any $k \in N$ , then

$3^{2n}+3n^2+7=9^{2k}+12k^{2}+7$

Now $9^{2k}<9^{2k}+12k^2+7 \leq (9k + 1)^2$ and the equality holds only at $k=1$ and the given expression can't be a perfect square for any $n>2$ . Thus, $3^{2n}+3n^2+7$ is a perfect square only for $n=2.1$

Hence $n=2$ is only solution for n, giving $\sum n=$ $\boxed{2}$ .