2
$\begingroup$

I have a polynomial $p$ with integer coefficients, which might have roots of multiplicity higher than one. I would like to obtain each root just once so I calculate $g = \gcd(p, p')$ and divide $q = p/g$. I know $g$ has integer coefficients, too, so my question is: does that division keep $q$ also in $\Bbb Z$?

I tried many random polynomials and it appears to be the case, but how can I be certain that this will always be the case? For example:

$$p = (x-7)\cdot(x-8)\cdot(x-8)\cdot(x-9)\cdot(x-9)\cdot(x-9)$$ $$p = x^6 - 50x^5 + 1040x^4 - 11518x^3 + 71631x^2 - 237168x + 326592$$

$$\gcd(p, p') = x^3 - 26x^2 + 225x - 648$$

$$p/q = x^3 - 24x^2 + 191x - 504$$

$\endgroup$
2
$\begingroup$

What you are asking is basically equivalent to showing that if $p$ and $g$ have integer coefficients, and $p = gq$, if it means $q$ has also integer coefficients. However, that need not be true:

$g = 2x + 6$, $q = \frac12x + 2$, $gq = p = x^2 + 7x + 12$

However if you impose that both $p$ and $g$ have their leading coefficient equal to $1$, then it must be the case that $q$ also has integer coefficients.

Should I prove it or do you want to attempt it? Start by proving that you need only consider the case when $g = x + a$ and then suppose that $q$ has non-integer coefficients and that $gq$ only has integer coefficients. What does that imply?

$\endgroup$
  • $\begingroup$ thank you. i attempt it, but probably will need help with the proof. i am just a amateur self-studied, so i might be slow. but i will try and let you know, thank you again. $\endgroup$ – Ondrej Stefik Nov 19 '16 at 0:01
  • 1
    $\begingroup$ @OndrejStefik being "slow" is fine. Not trying is not fine! Just give it a go. Start with small examples, if you must. Like $(x + a)(x + b)$. If $a $ is an integer, could $b $ be non-integer and still yield an integer-coefficient polynomial? And what for $(x + a)(x^2 + bx + c)$? Good luck and let me know how it went. $\endgroup$ – RGS Nov 19 '16 at 0:04
  • $\begingroup$ Thanks a lot for the answer but unfortunately I still don't quite understand it. Could you please provide an example, where p has (some) roots with multiplicity higher than 1 and p/q does not stay in Z? I'm asking because the given example of yours (p = x^2 + 7x + 12) does not have any multiple roots (its roots are -4 and -3). Because when p does not have any multiple roots its gcd(p, p') is 1 so p/q stays in Z $\endgroup$ – Ondrej Stefik Nov 20 '16 at 15:08

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.