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Working through A Book of Abstract Algebra, I encountered several exercises on roots of polynomials in [itex]Z_{n}[/itex] I was just wondering whether there exists something like a quadratic equation for polynomials of degree 2. If the solutions of the usual quadratic formula happen to be integers, can one simply take these modulo n to find solutions of the quadratic in [itex]Z_{n}[/itex]? What if they are not integers? Clearly a field extension is needed to find solutions, but how does this precisely work?
As an example, consider the equation [itex]x^{2} + x + 1 = 0[/itex] in [itex]Z_{2}[/itex]. The quadratic equation gives [itex]x = -\frac{1}{2} \pm \frac{1}{2} i[/itex]. Suppose you name the positive root as c. Then [itex]Z_{2}(c) = {0,1,c,1+c}[/itex] is the field extension. Now can one work with c in the same way as with [itex]x = -\frac{1}{2} \pm \frac{1}{2} i[/itex]? Clearly, you easily get contradictions like [itex]1 = -1 = 2c-1 = 3i = i[/itex] and so on.
As an example, consider the equation [itex]x^{2} + x + 1 = 0[/itex] in [itex]Z_{2}[/itex]. The quadratic equation gives [itex]x = -\frac{1}{2} \pm \frac{1}{2} i[/itex]. Suppose you name the positive root as c. Then [itex]Z_{2}(c) = {0,1,c,1+c}[/itex] is the field extension. Now can one work with c in the same way as with [itex]x = -\frac{1}{2} \pm \frac{1}{2} i[/itex]? Clearly, you easily get contradictions like [itex]1 = -1 = 2c-1 = 3i = i[/itex] and so on.