Baltic Way 2019

The answer is yes, and our example will be the polynomial $f(x)=x^4+2x$. Assume that there exist a constant $k$ such that $f(x)+k$ is reducible but has no rational zeros. This means that $f(x)+k$ factors as the product of two irreducible quadratic polynomials. Since $f(x)+k$ is monic so must the two factors be. Assume that $f(x)+k=(x^2+ax+b)(x^2+cx+d)$. By expanding, we find that $$x^4+2x+k=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$ By comparing coefficients we see that $a+c=0$, $b+d+ac=0$ and $ad+bc=2$. From the first equality we get $c=-a$, and the two others become $b+d=a^2$ and $(b-d)a=2$. From $b+d=a^2$ it follows that $b+d$ and $a$ has the same parity. Since $b+d$ and $b-d$ has the same parity, it follows that $2=(b-d)a$ is either odd or divisible by 4 which is a contradiction. Hence for any integer $k$ $f(x)+k$ is either irreducible or has a rational root, and the answer is therefore yes.