This question is similar to this other question:
Let $$ f(x,y):= \frac{x^y -1}{x+(-1)^y}$$ and $$ g(x,y):= \frac{f(x^{2y+1},y)}{f(x,y)}.$$
Let $y\ge1$ be an integer. Show that $g(x,y)$ is a polynomial function of $x$ with integer coefficients.
Moreover, it seems that that $g$ is irreducible in $\mathbb{Z}[x]$ if and only if both $y$ and $2y+1$ are prime numbers (in other words if and only if $y$ is a Sophie Germain prime).
Here are the first $g(x,y)$ for the integers $y$ such that $1\le y\le 4$: $$g(x,1)=1 $$ $$g(x,2)=1+x+x^2+x^3+x^4=\Phi_5(x)$$ $$g(x,3)=1-x+x^3-x^4+x^6-x^8+x^9-x^{11}+x^{12}=\Phi_{21}(x)$$ $$g(x,4)=\Big(1+x+x^2\Big) \Big(1-x^2+x^4\Big) \Big(1+x^3+x^6\Big) \Big(1-x^6+x^{12}\Big)=\Phi_{3}(x)\Phi_{9}(x)\Phi_{12}(x)\Phi_{36}(x)$$
It seems to me that $g(x,y)$ should be a product of cyclotomic polynomials $\Phi_k(x)$ where $k$ are divisors of $y(2y+1)$, but I could not derive a general formula.
If $y$ is even, then $$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}+1}}{\frac{x^y-1}{x+1}}=\frac{(x^{y(2y+1)}-1)(x+1)}{(x^{2y+1}+1)(x^y-1)}.$$ If $y$ is odd, then $$g(x,y)=\frac{\frac{x^{y(2y+1)}-1}{x^{2y+1}-1}}{\frac{x^y-1}{x-1}}=\frac{(x^{y(2y+1)}-1)(x-1)}{(x^{2y+1}-1)(x^y-1)}.$$ One has $$x^t-1=\prod_{d\mid t}\Phi_d(x),\qquad x^t+1=\prod_{\substack{d\mid 2t \\ d\nmid t}}\Phi_d(x).$$ As a result, for example, using that $\gcd(y,2y+1)=1$ you can show that for $y$ odd $$g(x,y)=\prod_{\substack{d\mid y(2y+1)\\ d\nmid y\text{ or }2y+1}}\Phi_d(x).$$ Can you do something similar when $y$ is even, and use these results to deduce exactly when $g(x,y)$ is irreducible?