Solution to Problem 5: "Perfection"

Doing this without a calculator is a bit of a hassle, but if you can do two-digit multiplication quickly, then it is manageable.

For reference, the four digit squares that begin with 3 are 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, and 3969.  Also, a square number can only end in 0, 1, 4, 5, 6, and 9. 

Row 4 contains a three-digit square that must be made up entirely of last digits of squares.  The only possibilities are 169, 400, 900, and 961.  Therefore, B4 is either 0 or 6.

If it were 0, then B would be 4900 and C would be 3600, so row 3 would be either 400 or 900 itself.  The only possibilities for A would be 3249 or 3844.  That means that row 2 would have to be either 896X or 296X.  We find that neither of these are squares, so we're stuck on this path.

Therefore, B4 is 6.  B is the square of a number that ends in 4 or 6 that is between 64 and 71, so the options for B is 4096 (64^2) and 4356 (66^2).  Of A and C, one is 3XX1 (which can be either 3481 or 3721) and the other is 3XX9 (which can be either 3249 or 3969).  Row 3 must be a square, and squares can't end in 8 or 2, so row 4 must be 169.  The only square 8XX is 841, and that's not row 3, so A is 3721.   Row 3 must be 256, so B is 4356 and C is 3969.  The only option for row 2 is 7396.  The completed diagram (and its square roots):

The three horizontal squares are 7396 256 169.

Comments: I didn't even try to do this one during the contest, and I'm surprised that fourteen people even tried.  If you've memorized the table of squares up to 100, it seems like it would be a snap.  I think it took me about fifteen minutes to solve it after the contest was over (still without a calculator), so I don't regret not trying it.