I'm looking for the smallest positive integer $n$ such that there's a quartic polynomial in $\Bbb Z_n [X]$ that has 8 distinct roots in $\Bbb Z_n$.

I have n equal to 15 with roots 1, 2, 4, 7, 8, 11, 13, 14 and equation $x^4-1=0$ but I'm unsure if you can go lower (is it possible to have n equal to 8?)

Also, while my math experience barely touches the surface of number theory and abstract algebra thus I'm not necessarily looking to prove this, I would like to understand why my claim of the minimality of n is true (or why yours for n less than 15 is) which I currently do not.

  • $\begingroup$ $8$ distinct roots? $\endgroup$ – Bernard Oct 7 '15 at 23:44
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    $\begingroup$ Yes; 8 distinct roots $\endgroup$ – Liam Cooney Oct 7 '15 at 23:44

Look at $4x(x+1)$, in $\mathbb{Z}_8$. Oops, a quadratic! Multiply by something.

For another example, this time monic, use $x(x+1)(x+2)(x+3)$, again in $\mathbb{Z}_8$.

  • $\begingroup$ Yeah, I was thinking in monic polynomials. $\endgroup$ – ajotatxe Oct 7 '15 at 23:52
  • $\begingroup$ I am literally minded! $\endgroup$ – André Nicolas Oct 7 '15 at 23:53
  • $\begingroup$ Golly gee I wonder what (this is as easy as I think right...?). Thank you! Damn easy to justify too ;) $\endgroup$ – Liam Cooney Oct 7 '15 at 23:58
  • $\begingroup$ Yes, pretty easy, of any two consecutive evens, one is divisible by $4$. $\endgroup$ – André Nicolas Oct 7 '15 at 23:59
  • $\begingroup$ I meant you can easily just multiply your cubic by x to get a quint of function that works $\endgroup$ – Liam Cooney Oct 8 '15 at 0:01

For monic examples, try $f(x) = x^4 + 2 x^3 + 7 x^2 + 6 x$ or $x^4 + 6 x^3 + 3 x^2 + 6 x$ in $\mathbb Z_8$.

EDIT: And at the next level, there are monic sextics over $\mathbb Z_{16}$ with $16$ different roots. For example, ${x}^{6}+3\,{x}^{5}+9\,{x}^{4}+5\,{x}^{3}+6\,{x}^{2}+8\,x$.

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    $\begingroup$ How are all you brilliant human beings coming up with these? Is there any real significance to it being monic or not monic? I'm just now learning what monic means... $\endgroup$ – Liam Cooney Oct 8 '15 at 0:00
  • $\begingroup$ Obviously it's easier to do in $\mathbb Z_8$ if all the coefficients are even. Multiplying everything by an odd number, on the other hand, doesn't change anything. $\endgroup$ – Robert Israel Oct 8 '15 at 0:39

Every number raied to $4$ ends in the same digit than the number itself, so $X^4-X$ has 10 roots in $\Bbb Z_{10}$, so $15$ is not the answer.


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