Asked by Chuck Bass
(x + y)^2 = x^2 + y^2
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Answered by
GPT-5 mini
AI
Expand the left side:
(x + y)^2 = x^2 + 2xy + y^2.
So the equation (x + y)^2 = x^2 + y^2 is equivalent to
2xy = 0.
Consequences (depending on the underlying number system):
- Over the real numbers (or any field of characteristic 0, e.g. Q, R, C): 2 ≠ 0 and there are no zero divisors, so 2xy = 0 implies xy = 0, hence x = 0 or y = 0.
- Over any field of characteristic 2: 2 = 0, so 2xy = 0 for all x,y and the equality holds for every pair (x,y).
- Over a general ring: the condition is simply 2xy = 0. That can hold with neither x nor y zero if the ring has zero divisors (for example in Z/6Z, x = 3, y = 1 gives 2·3·1 ≡ 0 (mod 6), so the equality holds though neither factor is 0).
Summary: The equality is equivalent to 2xy = 0. In familiar fields like R or C this means x = 0 or y = 0; in characteristic 2 it always holds; in rings with zero divisors there may be nontrivial solutions.
(x + y)^2 = x^2 + 2xy + y^2.
So the equation (x + y)^2 = x^2 + y^2 is equivalent to
2xy = 0.
Consequences (depending on the underlying number system):
- Over the real numbers (or any field of characteristic 0, e.g. Q, R, C): 2 ≠ 0 and there are no zero divisors, so 2xy = 0 implies xy = 0, hence x = 0 or y = 0.
- Over any field of characteristic 2: 2 = 0, so 2xy = 0 for all x,y and the equality holds for every pair (x,y).
- Over a general ring: the condition is simply 2xy = 0. That can hold with neither x nor y zero if the ring has zero divisors (for example in Z/6Z, x = 3, y = 1 gives 2·3·1 ≡ 0 (mod 6), so the equality holds though neither factor is 0).
Summary: The equality is equivalent to 2xy = 0. In familiar fields like R or C this means x = 0 or y = 0; in characteristic 2 it always holds; in rings with zero divisors there may be nontrivial solutions.
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