If `p(x)=(1+x^2+x^4++x^(2n-2))//(1+x+x^2++x^(n-1))` is a polomial in `x` , then find possible value of `ndot`

Explanation

To solve the problem, we need to analyze the polynomial \( p(x) \) given by:

\[ p(x) = \frac{1 + x^2 + x^4 + \ldots + x^{2n-2}}{1 + x + x^2 + \ldots + x^{n-1}} \]

**Step 1: Simplifying the numerator and denominator**

The numerator can be expressed as a geometric series:

\[ 1 + x^2 + x^4 + \ldots + x^{2n-2} = \frac{1 - x^{2n}}{1 - x^2} \]

The denominator can also be expressed as a geometric series:

\[ 1 + x + x^2 + \ldots + x^{n-1} = \frac{1 - x^n}{1 - x} \]

Thus, we can rewrite \( p(x) \):

\[ p(x) = \frac{\frac{1 - x^{2n}}{1 - x^2}}{\frac{1 - x^n}{1 - x}} = \frac{(1 - x^{2n})(1 - x)}{(1 - x^2)(1 - x^n)} \]

**Step 2: Further simplifying the expression**

Now we can simplify the expression:

\[ p(x) = \frac{(1 - x^{2n})(1 - x)}{(1 - x^2)(1 - x^n)} \]

**Step 3: Analyzing the conditions for \( p(x) \) to be a polynomial**

For \( p(x) \) to be a polynomial, the denominator must not introduce any factors that cancel with the numerator. This means we need to ensure that the roots of the denominator do not cancel out with the roots of the numerator.

**Step 4: Finding the roots**

The roots of the denominator \( 1 - x^n = 0 \) are the \( n \)-th roots of unity, which are given by:

\[ x = e^{\frac{2\pi i k}{n}} \quad (k = 0, 1, 2, \ldots, n-1) \]

The roots of the numerator \( 1 - x^{2n} = 0 \) are the \( 2n \)-th roots of unity, given by:

\[ x = e^{\frac{2\pi i m}{2n}} \quad (m = 0, 1, 2, \ldots, 2n-1) \]

**Step 5: Ensuring no cancellation**

For \( p(x) \) to be a polynomial, the roots of the denominator must not be roots of the numerator. This means that \( n \) must be such that the \( n \)-th roots of unity do not coincide with the \( 2n \)-th roots of unity.

This condition is satisfied when \( n \) is an odd number. If \( n \) is odd, the \( n \)-th roots of unity will not coincide with the \( 2n \)-th roots of unity.

**Conclusion: Possible values of \( n \)**

Thus, the possible value of \( n \) is:

\[ \text{n is odd} \]