?(x+y)?^2 = x^2 + y^2

^{2}= x

^{2}+ y

^{2}if the question marks are removed. If we are talking about integers: .. -2, -1, 0, 1, 2, ... then

^{2}+ 2xy + y

^{2}= x

^{2}+ y

^{2}, which would mean 2xy = 0, so x = 0 or y = 0. So the statement is true if either x or y is equal to zero, and otherwise it's false.

*which is equal to 0 on a 6-hour clock*. So on this 6-hour clock, if x = 2 and y = 3, then xy = 0 and we would have (x + y)

^{2}= x

^{2}+ y

^{2 }where neither x nor y was zero. So it really depends which number system you are working on whether an equality like this holds.