Exercise 1.3.31 ( No polynomial $f(x)$ of degree $1$ with integral coefficients can represent a prime for every positive integer $x$)

Question

Prove that no polynomial $f(x)$ of degree $>1$ with integral coefficients can represent a prime for every positive integer $x$.

Richard Ganaye · Accepted Answer

We may replace the hypothesis $\deg(f) >1$ by $\deg(f) \geq 1$.

\begin{proof} 
Let $f(x) = \sum_{i=0}^d a_i x^i$ a polynomial with integer coefficients $a_i$, where $\deg(f) \geq 1$.

If $f(j)$ is composite for every positive integer $j$, $f(x)$ doesn't represent any  prime.

Suppose now that there is some positive integer $j$ such that $f(j) = p$ is a prime number.
For all positive integers $j,k$,
\begin{align*}
f(j+kp) &= \sum_{i=0}^d a_i(j+kp)^i\
&\equiv \sum_{i=0}^d a_i j^i 
\pmod p\
\end{align*}
Therefore
$$ f(j+kp) \equiv f(j) \equiv 0 \pmod p.$$

Thus $p \mid f(j+kp)$ for all positive integers $k$.

Assume for contradiction that $f(j+kp) = p$ for all positive integers $k$. Then the polynomial $g(x) = f(x+kp) - p$ has infinitely many roots. Hence $g(x) = 0$, so $f(x) = p$. This is in contradiction with the hypothesis $\deg(f) \geq 1$. Therefore there is some positive integer $k$ such that
$$p \mid f(j+kp), \qquad p 
e f(j+kp).$$
 For this value of $k$, $f(j+kp)$ is composite. This proves that no polynomial $f(x)$ of degree $d \geq 1$ with integral coefficients can represent a prime for every positive integer $x$.
\end{proof}

Exercise 1.3.31 ( No polynomial $f(x)$ of degree $>1$ with integral coefficients can represent a prime for every positive integer $x$)

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Exercise 1.3.31 ( No polynomial $f(x)$ of degree $>1$ with integral coefficients can represent a prime for every positive integer $x$)

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