Plutog55
Give two different numbers, each of which is the square of the other!
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klimenkov
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Complex number or real?
pratu043
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interesting.
jazy
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1
pratu043
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two different numbers.
jazy
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Just realized that
klimenkov
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Solve[{x == y^2, y == x^2}, {x, y}]
{{x -> 0, y -> 0}, {x -> 1, y -> 1}, {x -> -(-1)^(1/3),
y -> (-1)^(2/3)}, {x -> (-1)^(2/3), y -> -(-1)^(1/3)}}
klimenkov
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Only in complex they will be different.
anonymous
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i will say \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and its conjugate
klimenkov
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@jazy = lazy
jazy
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I'm not at this level of math yet. I posted because I thought it to be a good answer, not because I was too lazy to help and do it myself. However I'll take it off it bothers. (: