Problem:
Solution:
For part (a), the second equation seems to be simpler and therefore we will start there. Assuming $ x \ne 0 $, we can write $ y = \frac{1}{x} $. Putting this back to the first equation gives:
$ x^2 + \frac{1}{x^2} - 1 = 0 $.
This works if $ x \ne 0 $, but what if $ x = 0 $? Turn out we do not need to worry about it because if $ x = 0 $, then the second equation cannot possibly be satisfied. Envisioning part (b), let's convert it into a polynomial by multiplying through $ x^2 $ and get:
$ x^4 - x^2 + 1 = 0 $
For part (b), we use the hint and get $ (xy - 1)(xy + 1) = x^2y^2 - 1 $. To eliminate $ y^2 $, we multiply the first equation by $ x^2 $ so we get $ x^2(x^2 + y^2 - 1) = x^4 + x^2y^2 - x^2 $.
Finally we get $ x^4 - x^2 + 1 = x^2(x^2 + y^2 - 1) - (xy - 1)(xy + 1) $, therefore $ x^4 - x^2 + 1 \in \langle x^2 + y^2 - 1, xy - 1 \rangle $
Solution:
For part (a), the second equation seems to be simpler and therefore we will start there. Assuming $ x \ne 0 $, we can write $ y = \frac{1}{x} $. Putting this back to the first equation gives:
$ x^2 + \frac{1}{x^2} - 1 = 0 $.
This works if $ x \ne 0 $, but what if $ x = 0 $? Turn out we do not need to worry about it because if $ x = 0 $, then the second equation cannot possibly be satisfied. Envisioning part (b), let's convert it into a polynomial by multiplying through $ x^2 $ and get:
$ x^4 - x^2 + 1 = 0 $
For part (b), we use the hint and get $ (xy - 1)(xy + 1) = x^2y^2 - 1 $. To eliminate $ y^2 $, we multiply the first equation by $ x^2 $ so we get $ x^2(x^2 + y^2 - 1) = x^4 + x^2y^2 - x^2 $.
Finally we get $ x^4 - x^2 + 1 = x^2(x^2 + y^2 - 1) - (xy - 1)(xy + 1) $, therefore $ x^4 - x^2 + 1 \in \langle x^2 + y^2 - 1, xy - 1 \rangle $
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