[amp_mcq option1=”2m – 1″ option2=”$$\frac{m}{n} – 1$$” option3=”$$\frac{{\left( {m + 1} \right)}}{2}$$” option4=”m – n – 1″ correct=”option1″]
The correct answer is $\boxed{\frac{{\left( {m + 1} \right)}}{2}}$.
A primitive polynomial of degree $m$ is an irreducible polynomial in such a way that it is a factor of $x^n + 1$, where $n = \frac{{\left( {m + 1} \right)}}{2}$.
Here is a brief explanation of each option:
- Option A: $2m – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $2 \cdot 3 – 1 = 3$.
- Option B: $\frac{m}{n} – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $\frac{3}{2} – 1 = \frac{1}{2}$.
- Option C: $\frac{{\left( {m + 1} \right)}}{2}$ is always a factor of $x^n + 1$. This is because $x^n + 1$ can be factored as $(x + 1)(x^{n – 1} – x^{n – 2} + \cdots + 1)$. The term $x + 1$ is always a factor of $x^n + 1$, and the other terms in the product are all divisible by $2$. Therefore, $x^n + 1$ is always divisible by $\frac{{\left( {m + 1} \right)}}{2}$.
- Option D: $m – n – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $3 – 2 – 1 = 0$.
Therefore, the correct answer is $\boxed{\frac{{\left( {m + 1} \right)}}{2}}$.