Let $f\in\mathbb{Z}[X]$ be an irreducible polynomial. Is there an integer $a\neq 0$ such that $f(X)+a$ is also irreducible in $\mathbb{Z}[X]$?


Can this be also extended to $\mathbb{Q}[X]$?



Is there an integer a such that f(X)+a is irreducible in Z[X]?
Yes, and you don't need $f$ irreducible. The following irreducibility criterion suffices and shows that infinitely many $a$ work. 


**Lemma:** Let $g(x) = a\_n x^n + ... + a\_0 \in \mathbb{Z}[x]$ be such that $a\_0$ is prime and 


$$|a\_0| > |a\_1| + ... + |a\_n|.$$


Then $g(x)$ is irreducible.


*Proof.* The condition on the coefficients ensures that all complex roots of $g$ have absolute value greater than $1$. But if $g$ is reducible then at least one of its irreducible factors has constant term $\pm 1$, hence at least one of its roots has absolute value less than or equal to $1$; contradiction. $\Box$


