, 2013

We rewrite the integers from the list in terms of their prime factorizations: $$1,2^1,3^1, 2^2, 5^1, 2^1 3^1, 7^1, 2^3, 3^2$$ A positive integer larger than one is a perfect square if and only if each of its prime factors occurs an even number of times.

Since the integers in the list above contain in total only one factor of 5 and one factor of 7, then neither 5 nor 7 can be chosen to form a product that is a perfect square.

This leaves us with seven integers $1,2^1,3^1, 2^2, 2^1 3^1, 2^3, 3^2$, from which we need to choose six.

When the seven given integers are multiplied together, their product is $2^{1+2+1+3}3^{1+1+2}=2^7 3^4$.

We can think of choosing six of the seven numbers and multiplying them together as choosing all seven and then dividing out the one we did not choose.

To divide the product $2^7 3^4$ by one of the integers to obtain a perfect square, the divisor must include an odd number of factors of 2 (since the product of all seven includes an odd number of factors of 2) and an even number of factors of 3 (since the product includes an even number of factors of 3). (Note that “an even number of factors of 3" includes the possibility of zero factors of 3.)

There are two such numbers in the list: $2^1$ and $2^3$.

(Alternatively, we could have divided the product by each of the seven numbers to determine which results in a perfect square.)

Therefore, the two sets of six numbers that satisfy the given conditions should be $1,3^1, 2^2, 2^1 3^1, 2^3, 3^2$ (whose product is $2^6 3^4$) and $1,2^1,3^1, 2^2, 2^1 3^1, 3^2$ (whose product is $2^4 3^4$).

Thus, we can set $m^2 = 2^6 3^4$, which gives $m=2^3 3^2=72$, and $n^2 = 2^4 3^4$, which gives $n=2^2 3^2 = 36$.

Finally, $m+n=72+36=108$.