Question

Find the gcd of x2+6x+1 and x2+3 in Z7[x].

Answer 1

Answer on Question #44364 – Math - Abstract Algebra

Problem

Find the gcd of $x^2 + 6x + 1$ and $x^2 + 3$ in Z7[x].

Solution

$$\begin{array}{l} x^2 + 6x + 1 = (x^2 + 3) \cdot 1 + (6x - 2) \\ \operatorname{gcd}(x^2 + 6x + 1, x^2 + 3) = \operatorname{gcd}(6x - 2, x^2 + 3) \\ x^2 + 3 \equiv (6x - 2)(6x + 2) + 0 \text{ (in } \mathbb{Z}_7[x]). \end{array}$$ $$\operatorname{gcd}(6x - 2, x^2 + 3) = 6x - 2$$

Answer

$$\operatorname{gcd}(x^2 + 6x + 1, x^2 + 3) = 6x - 2$$

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