The following is an exercise in Victor V. Prasolov's Polynomials (Second edition, Page no. $74$, exercise $2.10$)
Problem: Let $p>3$ be a prime and $n<2p$ a natural number. Prove that $$P(X)=X^{2p}+pX^n-1$$ is irreducible over $\mathbb{Z}[X]$.
My approach: Let, if possible, assume that $P=fg$ for $2$ polynomials $f,g\in\mathbb{Z}[X]$ with $\deg(f),\deg(g)\geq1$. For a polynomial $q\in\mathbb{Z}[X]$ let $\overline{q}$ denote its reduction modulo $p$. Then $$\overline{P}=X^{2p}-1=(X^p+1)(X^p-1)=(X+1)^p(X-1)^p$$ Therefore $\overline{f}\overline{g}=(X-1)^p(X+1)^p$.
Edit: According to @WhatsUp's solution we must have $\overline{f}=X^p+1,\overline{g}=X^p-1$(or vice-versa).
Update: Then we have $f(X)=X^p+1+pF(X)$ and $g(X)=X^p-1+pG(X)$ for some $F,G\in\mathbb{Z}[X]$. Since $\deg(f)+\deg(g)=\deg(P)=2p$ and $P$ is monic and $X^n$ has coefficient $p$ and all other coefficients are $0$ we must have $\deg(F),\deg(G)\leq p-1$. If $F,G$ were both non-zero then equating $P$ and $fg$ we have $$(G(X)-F(X))+X^p(G(X)+F(X))+pG(X)F(X)=X^n$$ Therefore, since $p>3$ is an odd prime, evaluating at $X=1$ we get $$G(1)(2+pF(1))=1$$ This means $$pF(1)=\pm1-2$$ equivalently, $pF(1)=-3$ or $pF(1)=-1$. The latter is not possible. For the former case we must have $p=3$ but $p>3$ is given.
A contradiction in all cases since $G(1),f(1)\in\mathbb{Z}$ and $p>3$ is an odd prime!
Then one of $F,G$ must be the $0$ polynomial. Let $F\equiv0$. Let $G$ be a non-zero polynomial. Then we have $$X^{2p}+pX^n-1=(X^p+1)(X^p-1+pG(X))=X^{2p}-1+pX^pG(X)+pG(X)$$ That means $$pX^pG(X)-pG(X)=pX^n\implies X^pG(X)-G(X)=X^p\implies G(X)(X^p-1)=X^n$$ which implies $0=G(1)(1^p-1)=1^n=1$, a contradiction! Therefore such $F,G$ can't exist. Again $(X^p+1)(X^p-1)\neq P(X)$. Then $P$ can't be factorized over $\mathbb{Z}[X]$ or equivalently $P$ is irreducible over $\mathbb{Z}[X]$.
Is this argument correct? Can someone tell me if this proof is correct or wrong?
To justify your argument, I will show that we must have $\overline f = X^p + 1$, $\overline g = X^p - 1$ or vice-versa.
Suppose that this is not the case. Then either $X + 1$ or $X - 1$ would divide both $\overline f$ and $\overline g$.
Let's say $X - 1$ divides both of them (the other case being similar).
This means that $p\mid f(1)$ and $p\mid g(1)$, hence $p^2 \mid f(1)g(1) = P(1)$.
But we have $P(1) = p$, a contradiction.
To continue the argument, let $F, G$ be as in your post. Thus we have $$X^{2p} + pX^n - 1 = (X^p + 1 + pF)(X^p - 1 + pG).$$ This leads to $$(X^p - 1)F + (X^p + 1)G + pFG = X^n.$$
Now plug in $X = 1$, we get $(2 + pF(1))G(1) = 1$, which means $2 + pF(1)$ must be $1$ or $-1$.
Hence $pF(1)$ must be $-1$ or $-3$. This is only possible for $p = 3$, which we check manually.