Hmmm, that equation isn’t really false, is it? 
But, then again, realizing why that equation isn’t false reveals the answer to the puzzle. :smack:
Am I being whooshed or is there some underlying truth to 1 = -1?
On further inspection of flex727’s first line I assume it comes from the derivation
Sqr(x-y) = Sqr(x-y)
Sqr(x-y) = Sqr[(-1) * (y-x)]
Sqr(x-y) = Sqr(-1) * Sqr(y-x)
Sqr(x-y) = i * Sqr(y-x).
I can see that the last statement is false, as I saw it was at the beginning of flex727’s problem. It’s still early out here but I can’t seem to find the error in that derivation. Math geeks?
I think it has to do with the square root function being multi-valued.
1 (like all numbers) really has two “square roots”: 1 and -1: (-1)[sup]2[/sup]=1. It’s often convention for the “sqrt” function to imply the positive root, but messing about too haphazardly with this results in ridiculousness like that above.
Similarly, -1 has two square roots: i and -i.
So, 1[sup]1/2[/sup] = i * (-1)[sup]1/2[/sup] is extremely dangerous to write, as both sides of that equation have two values. 1[sup]1/2[/sup] has the values 1 and -1; i * (-1)[sup]1/2[/sup] has the values ii = -1 and i-i = 1.
Pick and choose which values you want, and the relation can be either true or false. I’m pressed for time so I can’t flesh this out as thoroughly as I’d prefer to, but I think the whole “proof” makes use of this sort of logical nonsense.
[Lurker mode off]
‘Sqr(x-y) = i * Sqr(y — x)’ is not the problem.
Let’s rename a couple of the variables in the puzzle:
Start from this obvious truth: Sqr(x-y) = i * Sqr(y — x).
Since this is plainly the case for any numbers x and y, put x = a and y = b. Then
Sqr(a — b) = i * Sqr(b — a).
On the other hand, since that obvious truth I started with is the case for any numbers x and y, I could equally well put x =** c** and y = d, to get another, equally true statement:
Sqr(c — d) = i * Sqr(d — c).
Now, if P = Q and R = S, then obviously P * R = Q * S. So:
Sqr(a — b) * Sqr(c — d) = i [sup]2[/sup] * Sqr(b — a) * Sqr(d — c).
Replace a,b,c,d with x,y and we get:
x — y = i[sup]2[/sup] * (y — x)
<=>
x — y = x — y
[Lurker mode on]
b4i4q ru/18 qt pi? 
Unfortunately I can’t see some of the binary operations (in flex’s or moigle’s post (i.e. they’re not in my character set),
what is the binary operation, that I’ve replaced with the %
“sqrt(x-y) = i*sqrt(x % y)”?
Am I right in assuming it’s a minus sign (though why can I see some minus signs and not others)?
Though it’s hard to follow not seeing all the binary operations, it looks like a variation on the one that Richard posted.