2

Consider polynomials of degree two over $\mathbb Z$:

$f = ax^2+bx+c$

The discriminant is

$D = b^2-4ac$

And we can show that $D=2$ is not a possible value for $D$. I wonder if the value $D=2$ occurs, if we just extend our set of polymonials to degree-3 (4, 5, 6, ...) polynomials over $\mathbb Z$?

Indeed, for degree 3:

$f = a_3x^3+\ldots+a_0$, $D=a_1^2a_2^2-4a_0a_2^3-4a_1^3a_3+18a_0a_1a_2a_3-27a_0^2a_3^2 $

I cannot find a polynomial with $D=2$ for $a_i\in [-20,20]$.

My question is: Does the discriminant of a polynomial with variable degree assume every value in $\mathbb N$ (or $\mathbb Z$)?

Community
  • 1
Dan
  • 744
  • 3
  • 13
  • $b^2-4ac$ cannot be a number of the form $4k+3$! – Jack D'Aurizio Apr 27 '15 at 15:39
  • 6
    It is a theorem of Stickelberger that any discriminant is $0$ or $1 \bmod 4$. And not all values satisfying one of those congruence conditions are necessarily possible, e.g., no polynomial in $\mathbf Z[x]$ of degree at least $2$ has discriminant $1$. A related question: http://math.stackexchange.com/questions/820990/is-every-nonzero-integer-the-discriminant-of-some-algebraic-number-field. – KCd Apr 27 '15 at 15:40
  • Now $f$ could be reducible, hence we do not have $disc(f)=\Delta_K$ for a number field $K$. Then why is $f$ still not $2 \pmod 4$? – Dan Apr 28 '15 at 09:45

0 Answers0