0
$\begingroup$

Suppose $f(x)$ is an integer polynomial that can be factored as $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ are both integer polynomials of positive degree. Prove that there exists an integer $n$ such that the integer $|f(n)|$ is composite.

We can see that $|f(x)| = |g(x)|\cdot|h(x)|$. Since $n \in \mathbb{N}$, $|g(n)|$ and $|h(n)|$ are integers.

But how can we show that $|g(n)|\cdot|h(n)|$ is not prime?

$\endgroup$
0
$\begingroup$

Hint: a non-constant polynomial can take any given value at most finitely many times.

Therefore $G = g^{-1}\left(\,\{-1,0,1\}\,\right)$ and $H = h^{-1}\left(\,\{-1,0,1\}\,\right)$ are both finite. Then, for any $n \in \mathbb{Z} \setminus (G \cup H)\,$ $f(n)$ will be a composite number (why?).

$\endgroup$
  • $\begingroup$ im not entirely sure what you mean, could you elaborate a little further please $\endgroup$ – u123435 Mar 31 '17 at 3:14
  • $\begingroup$ $g(n)$ is an integer, and $g(n) \ne 0,\pm 1$ since $n \not \in G\,$. Same goes for $h(n)\,$. Then $f(n)=g(n)h(n)$ is a product of two integers, neither of which is $0$ or $\pm 1$. $\endgroup$ – dxiv Mar 31 '17 at 3:16
0
$\begingroup$

Note $p > 0$ the degree of $g$ and $q > 0$ the degree of $h$. Suppose that for all $n$, $| f (n) |$ is prime. Since $| g (n) |$ and $| h (n) |$ divide $| f (n) |$, we have for all $n$, $| g (n) | = 1$ or $| h (n) | = 1$.

Let $G = \{ n \in \mathbb{Z} \mid | g (n) | = 1 \}$ and $H = \{ n \in \mathbb{Z} \mid | h (n) | = 1 \}$. From the above, one of these two sets must be infinite.

  • Suppose that $G$ is infinite, and let $G^+$ (resp. $G^-$) be the subset of $G$ such that $g (n) = 1$ (resp. $g (n) = -1$). We have $G = G^+ \cup G^-$, so at least one of $G^+$ or $G^-$ must be infinite. From this we deduce that the polynomial $g$ has infinitely many (complex) roots, and is therefore constant. This is a contradiction with $p > 0$.

  • If $G$ is finite, then $H$ must be infinite, and a similar reasoning leads to $h$ being constant, which contradicts $q > 0$.

Therefore our hypothesis that $| f (n) |$ is prime for all $n$ is false, and thus there exists some $n_0$ such that $| f (n_0) |$ is composite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.