Is the cubic polynomial $x^3-ax-(2a+1)$ irreducible over $\mathbb Z$ for all positive integers $a$?
one way to prove irreducibility is to use eisenstein's criterion. we want to find prime $p$ st:
$p\mid a$
$p\mid 2a+1$
$p^2\nmid 2a+1$
However, the first 2 conditions imply $p\mid 1$, a contradiction. so this approach fails.
another approach is to use the rational root thm.
let $x^3-ax-(2a+1)=(x+b)(x^2+cx+d)=x^3+(b+c)x^2+(bc+d)x+bd$
where b,c,d integers
$b+c=0$
$bc+d=-a$
$bd=-(2a+1)$
there was a link to a similar problem, now deleted.
If a cubic is reducible, then it must have a linear factor $x+b$, so $-b$ is an integer root.
$$-b^3+ab-(2a+1)=0$$ $$a(b-2)=b^3+1$$ $$a=\frac{b^3+1}{b-2}$$ $$(b-2)\mid 9$$ \begin{align}b-2&=-9,-3,-1,1,3,9\\b&=-7,-1,1,3,5,11\\b&\neq -1,1\end{align} so $(a,b,c)=(38,-7,11),(28,3,-19),(42,5,-17),(148,11,-27)$.
\begin{align}(x-7)(x^2+7x+11)&=x^3-38x-77\\ (x+3)(x^2-3x-19)&=x^3-28x-57\\ (x+5)(x^2-5x-17)&=x^3-42x-85\\ (x+11)(x^2-11x-27)&=x^3-148x-297\end{align}
so there are 4 $a$'s such that the cubic is reducible. Simply stating (but not deriving) one of the 4 factorizations is enough to disprove the question.
thanks to @dxiv for the suggestion.