There are two regular polygons: one with X sides and one with Y sides with Y>X. Also, coincidentally, the latter has internal angles that are each (Y-X) degrees greater than the former. Also, X+Y is a perfect square.

What are X and Y?

(In reply to solution by Dej Mar)

Nicely laid out, Dej Mar. I came to the same conclusion by what appears to be the same reasoning. I wondered afterwards, though, whether the step of checking every factorization was necessary, and it seems not.

The factors of two cannot be split between X and Y. If they were, one would be = 2 mod 4 and the other = 0 mod 4 so their sum would = 2 mod 4. But all perfect squares are = 0 mod 4 or 1 mod 4 so such a sum can't be a perfect square. One of X and Y is odd, then, and the other is a multiple of 8.

The factors of 3 similarly cannot be split. If they were, the sum would be a multiple of 3 and hence need to be a multiple of 9. That would require that X/3 + Y/3 = 0 mod 3 to "create" a second multiple of 3. But the only factors available are 8 (by the first point above) and 5 which are both = -1 mod 3. Whether these factors are part of the same X or Y or split, X/3 = Y/3 != 0 mod 3 and so the sum is not a mulitple of 9. Hence, one of X and Y is a multiple of 9 and the other is not a multiple of 3.

This means that there are only three factors to distribute to X and Y: 8, 9, and 5. Now, one of X and Y is a multiple of 5 (whichever gets the 5 factor) but the sum isn't. Perfect squares that aren't multiples of 5 must be = +/-1 mod 5. 8 = -2 mod 5 and 9 = -1 mod 5 so if the 8 factor is on the opposite X/Y from the 5, the sum can't be a perfect square--it's either +2/-2 mod 5 depending on which factor gets the 9. So one factor has both the 8 and 5 and is therefore 40. Since 40 > sqrt(360), Y=40 and X=9.

(I've neglected to consider the case where Y=360 and X=1 above; although this meets the requirements that XY=360 and X+Y=361=19², It fails to describe two regular polygons--such entities require X,Y >= 3)

There isn't anything wrong with iterating through the possible factorizations (in truth, that's how I arrived at the solution at first) but I think it's interesting to explore whether a solution can be derived as well, and in this case, it can.

Posted by Paul on 2008-10-29 15:36:43