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 $p(1) \ne p(-1) \ne  0$ ?
 A: The simpler counter-example is obviously $P(X)=-X+a_0$.
Whatever the nature of $a_0$ it is clear that $P(a_0)=0$.

If we have a polynomial $P(X)=a_nX^n+...+a_1X+a_0=a_n(X-x_1)(X-x_2)...(X-x_n)$
Then the Newton's identities give a relation between the coefficients and the roots of the polynomial (we assume $a_n\neq 0$).
$\begin{cases}
\sigma_0=1\\
\sigma_1=x_1+x_2+...+x_n=\sum\limits_{i=1}^nx_i\\
\sigma_2=x_1x_2+x_1x_3+...+x_{n-1}x_n=\sum\limits_{1\le i,j\le n}x_ix_j\\
...\\
\sigma_k=\sum\limits_{1\le i_1,...,i_k\le n}x_{i_1}x_{i_2}...x_{i_k}\\
...\\
\sigma_n=x_1x_2...x_n=\prod\limits_{i=1}^nx_i\\
\end{cases}$
Then $\displaystyle \sigma_k=(-1)^k\frac{a_{n-k}}{a_n}$
In particular in the problem that interests us at the moment we have $\sigma_n=\prod\limits_{i=1}^n x_i=(-1)^n\frac{a_0}{a_n}$
The constraint that $x_1=a_0$ for instance only results in $\prod\limits_{i=2}^n x_i=\frac{(-1)^n}{a_n}$ which is a very loose constraint for a product of $n-1$ algebraic numbers. 
For instance I can take $x_2=x_3=...=x_{n-1}=1$ and $x_n=\frac{(-1)^n}{a_n}$ which is rationnal and thus algebraic, but of course you can guess that other choices for the $x_i$ are possible (and in infinite number).


Maybe you were thinking of a different case where $a_n=1$ and all roots are also integers, but even in this case there is a solution.
$\prod\limits_{i=1}^n x_i=(-1)^na_0$ by Gauss theorem this implies that one root is $\pm a_0$ and the others are all $\pm 1$
Thus all polynomials of the form $P(X)=(X-a_0)(X-1)^m(X+1)^n$ with $a_0\text{ prime}$ and $m,n$ positive or null integers verifying $m+n$ odd would have $P(a_0)=0$.
A: For $q$ being any number, $p(x)=x^n-(q^{n-1}+1)x+q$ has a zero at $x=q$.
A: Counter example:
$$
 p(x) = (x-3)(x-1) = x^{2}-4x+3
$$
where the root are, as noted by @hardmath, $\left( 3, 1 \right)$.
A: The conjecture is far from being true.  A good point of comparison is the Rational Root Test.  For a polynomial with integer coefficients, say $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$, and a rational root $\dfrac{r}{s}$ we must have $r|a_0$ and $s|a_n$
A: Consider the polynomial $$f(x)=x^2 - 6 x + 5$$
The constant term here is $5$, which is prime.
However, \begin{align}f(5)&=5^2-6\times 5+5\\
&=25-30+5\\
&=0\end{align}
And thus $5$ is a root of the equation.
Therefore, I have disproved your hypothesis through contradiction

For higher degree polynomials, anything of the form $$f(x)=(x\pm p)(x+1)^a(x-1)^b$$ for $p$ prime, and $a,b$ integers will form a contradiction.
Examples include $$x^4 + 8 x^3 + 6 x^2 - 8 x - 7 = (x+7)(x+1)^2(x-1)$$ which has $p=-7$, $a=2$, $b=1$
and $$x^6 - 14 x^5 + 11 x^4 + 28 x^3 - 25 x^2 - 14 x + 13 = (x-13)(x+1)^2(x-1)^3$$ which has $p=13$, $a=2$, $b=3$

To add the constraint that $f(1)\neq f(-1)\neq 0$, then we can choose $$x^3 - 11 x^2 + x - 11=(x-11)(x^2+1)$$ 
Here we have $f(1)=-20$, $f(-1)=-24$ but $f(11)=0$
As long as the second polynomial is an irreducible one, with a constant of $1$, then this will form a contradiction
