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No, this is true, proof by induction:
(x+y)^2 = x^2 + y^2
For (x,y) = (0,0) :
(x+y)^2 = (0+0)^2 = (0)^2 = 0
x^2 + y^2 = 0^2 + 0^2 = 0 + 0 = 0
Q. E. D.
Thus it should follow that it applies to every other case. (The other cases are left as an exercise for the reader.)
Go home, ChatGPT, you’re drunk.
From the first one you get (a = -1/sqrt(2), b=1/sqrt(2)) that 0=1. Thus the field of real numbers is trivial. It's thus not surprising that every function into it is linear lol.