Question

Find unit set of Z√m. Where m is not perfect square.

Answer 1

Answer on Question #82164 – Math – Abstract Algebra

Question

Find unit set of $Z[\sqrt{m}]$, where $m$ is not perfect square.

Solution

We have to find all integer solutions $(x, y)$ to the Pell equation (1):

$$a = x + y\sqrt{m};$$ $$b = x - y\sqrt{m};$$ $$ab = 1;$$ $$(x + y\sqrt{m})(x - y\sqrt{m}) = 1.$$ $$x^2 - m y^2 = 1.$$

1) $m = -1$:

$$x^2 + y^2 = 1.$$

This equation has four trivial solutions: $(\pm 1, 0)$, $(0, \pm 1)$ and 4 units:

$$-1; 1; -i; i.$$

2) $m < -1$, $m \in \mathbb{Z}$:

$$x^2 + (-m)y^2 = 1.$$

In this case we have two trivial solutions: $(\pm 1, 0)$ and 2 units.

$$-1; 1.$$

3) In case of the positive non-perfect number for $m$ we have two trivial solutions: $(\pm 1, 0)$, 2 units: -1; 1, and infinite number of non-trivial solutions:

$$(x; y) = (\pm x_n; \pm y_n), \text{ where:}$$ $$x_n + y_n\sqrt{m} = (x_0 + y_0\sqrt{m})^n, \ n \in \mathbb{N}.$$

For example, if $m = 2, 3, 5$:

a) $m = 2$:

$$x^2 - 2y^2 = 1:$$ $$x_n + y_n\sqrt{2} = (3 + 2\sqrt{2})^n.$$

b) $m = 3$;

$$x^2 - 3y^2 = 1;$$ $$x_n + y_n\sqrt{3} = (2 + \sqrt{3})^n.$$

c) $m = 5$;

$$x^2 - 5y^2 = 1;$$ $$x_n + y_n\sqrt{5} = (9 + 4\sqrt{5})^n.$$

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