1
$\begingroup$

Consider two non constant polynomials, $f(x)$ and $g(x)$, with coefficients in Z. if $$f(x)g(x)-2015=0$$ has at least 33 distinct solutions in Z, then show that both f and g are polynomials of degree at least 3.

$\endgroup$

closed as off-topic by Ian Miller, Shailesh, John B, Ben Sheller, Watson Apr 1 '16 at 8:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ian Miller, Shailesh, John B, Ben Sheller, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Please indicate what you have tried and where you are stuck. This will help people better tailor their answer to your background and situation. It will also demonstrate that you are interested in your question and not just looking for someone to do your homework for you - Math.SE is not a homework site. $\endgroup$ – Ian Miller Mar 31 '16 at 23:44
  • $\begingroup$ What are your own thoughts? For instance, have you thought about what they are actually asking about here? Obviously if there are $33$ solutions, $f(x)g(x)$ must have quite a high degree (like, at least $33$). Why are they asking about as low a number as $3$? $\endgroup$ – Arthur Mar 31 '16 at 23:44
  • $\begingroup$ @HenryW. non-constant ${}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Mar 31 '16 at 23:48
  • 2
    $\begingroup$ @Mirko : if you allow $g(x) = 1$ then $f(x) = 2015+ \prod_{k=0}^n (x-k)$ is a solution (take whatever you need for $n$, 32 in this case) $\endgroup$ – Tryss Mar 31 '16 at 23:55
  • 1
    $\begingroup$ @Tryss: this is why the problem says non-constant. Presumably if we let $n=32$ in your expression we get a polynomial that is irreducible over the integers and this remains true no matter what collection of $k$'s we take. We are to prove that. $\endgroup$ – Ross Millikan Apr 1 '16 at 0:03
6
$\begingroup$

This is the general idea :

$2015 = 5\times 13 \times 31$

So as the polynomials are at coefficient in $\Bbb Z$, g and f will take values in this set at each root of the polynomial $P$:

$\{ \pm 1, \pm 5, \pm 13, \pm 31 ,\pm 65, \pm 155, \pm 403, \pm 2015 \}$

And every value of $g$ at a root is "conjugate" with another value of $f$ at this root (-1 goes with -2015, 31 goes with 65, etc.).

But this gives you only 16 distinct roots if your polynomial is of degree 1 (as f and g can take only once each of these values), and 32 roots if your polynomial is of degree 2, so you need at least degree 3

$\endgroup$
  • $\begingroup$ Oh, so sort of using Pigeon Hall principle? $\endgroup$ – mentorship Apr 1 '16 at 0:06
  • $\begingroup$ @mentorship : yes, it seems to works here $\endgroup$ – Tryss Apr 1 '16 at 0:07
  • $\begingroup$ @mentorship who is Pigeon Hall? I thought you spell her Pigeon Whole... $\endgroup$ – Mirko Apr 1 '16 at 0:09
  • 1
    $\begingroup$ it's actually pidgeon-hole $\endgroup$ – Jorge Fernández Hidalgo Apr 1 '16 at 0:09
  • 1
    $\begingroup$ You are missing the divisor $155$, which makes it exactly $32$ :) $\endgroup$ – Jorge Fernández Hidalgo Apr 1 '16 at 0:23

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