Let $p>3$ be a prime number. Prove that there doesn't exist a pair of polynomials $(f,g)\in{\mathbb{Z}[X]\times\mathbb{Z}[X]}$ such that:
$X^{2p}+pX^{p+1}-1=[(X+1)^p+p\cdot f(X)]\cdot[(X-1)^p+p\cdot g(X)]$
My best attempt at this problem was to look at the roots of the polynomial from the left hand side and prove that it has no real roots (now I know that it does). If I would have proven this, then I would have known that over $\mathbb{R}$, the polynomial is composed of irreducible components of degree 2 (with roots $z$ and its complex conjugate).
For the sake of contradiction assume such factorization, then at $X=1$ we have $p=(2^p+pf(1))\cdot (p\cdot g(1))$ and so especially $1=(2^p+pf(1))\cdot g(1)$. Since $f(1),g(1)$ are integers, $2^p+pf(1)=\pm 1$ and modulo $p$ we get $2^p \equiv \pm 1 \pmod{p}$. By Fermat's little theorem we know $2^p \equiv 2 \pmod p$ and so $2 \equiv \pm 1 \pmod {p}$. Thus $p \mid 3$ or $p \mid 1$, both impossible for $p>3$.