For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial. 
For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.



Let $p$ be a polynomial with constant term $a_0$ and if $a_0$ is prime then $p(a_0) \ne  0$ 


I just thought of this while working on some other problem.
Is this true ?
My attempt :- 
$$|a_0/a_n| = |r_1 ... r_n|$$
If, $r_1 = a_0$
$$1 = |a_n||r_2 ... r_n|$$
Also $$|a_1/a_n| = |r_2...r_n + r_1r_3...r_n + ... + r_1...r_{n-1}|$$
Or $$|a_1| =  |r_1|\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$$
I am having difficulty in proving that $\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$ not a integer if $|r_1 ... r_n|^{-1}$ is a integer.
Any hints ? 

Ok this is false but can somebody prove/disprove this if $p(1) \ne p(-1) \ne  0$ ?

Counter example:
$$
 p(x) = (x-3)(x-1) = x^{2}-4x+3
$$
where the root are, as noted by @hardmath, $\left( 3, 1 \right)$.
