Prove $f(x)$ has no rational number roots

Question

Let $f(x)=ax^2+bx+c, \ a,b,c\in \mathbb{Z}$, and such $$|f(1)|,|f(2)|,|f(3)|,|f(4)|,|f(5)|$$ are prime numbers,

show that $f(x)=0$ has no rational number roots

Answer 1

Hints: If $f(x) = ax^2+bx+c$ has a rational root, then both of its roots are rational, and it can be factored as $f(x) = (p_1x+q_1)(p_2x+q_2)$ for some integers $p_1,q_1,p_2,q_2$.

If $|f(x)| = |p_1x+q_1| \cdot |p_2x+q_2|$ is prime for some integer $x$, then either $p_1x+q_1 = \pm 1$ or $p_2x+q_2 = \pm 1$.

For how many distinct values of $x$ can we have $p_1x+q_1 = \pm 1$ or $p_2x+q_2 = \pm 1$?