If $\mathrm{p}\left(\mathrm{x}\right)=\frac{1+{\mathrm{x}}^2+{\mathrm{x}}^4+\cdots +{\mathrm{x}}^{2\mathrm{n}-2}}{1+\mathrm{x}+{\mathrm{x}}^2+\cdots +{\mathrm{x}}^{\mathrm{n}-1}}$ is a polynomial in x, then n can be

$\mathrm{p}\left(\mathrm{x}\right)=\left(\frac{1-{\mathrm{x}}^{2\mathrm{n}}}{1-{\mathrm{x}}^2}\right)\left(\frac{1-\mathrm{x}}{1-{\mathrm{x}}^{\mathrm{n}}}\right)=\frac{1+{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}$

As p(x) is a polynomial, x = -1 must be a zero of 1 + xⁿ. Hence, 1 + (-1)ⁿ = 0. So, n must be odd.