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Take a random polynomial $f$ with integer coefficients (e.g., choose coefficients between $1$ to $B$ of a fixed degree $n$ and let $B$ tend to $\infty$). Using computer we noted that the 2-valuation of the polynomial tends to be even, and much larger than one may naively expected.

Are there results about the 2-valuation of the discriminant (other than the classical result that it $\equiv 1, 0 \mod 4$)?

Here is some data for degree = 10, B=1000, we chose 100,000 random polynomials and the following is a list of 2-valuation: [50156, 0, 12231, 4191, 10462, 2551, 5974, 1034, 4435, 1105, 3107, 984, 1283, 447, 672, 337, 356, 150, 171, 88, 94, 41, 43, 20, 28, 14, 5, 6, 3, 3, 4, 1, 2, 1, 1]

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    $\begingroup$ These properties seem to hold already in the toy example $n=2$. Namely, let $\Delta = b^2-4ac$. If $v_2(b^2) > v_2(4ac)$, then necessarily the 2-valuation is even. If $v_2(b^2) < v_2(4ac)$, the 2-valuation is even with probability $\frac{5}{9}$. $\endgroup$
    – Ofir Gorodetsky
    Commented Sep 25, 2016 at 11:44
  • $\begingroup$ What would be the "naive expectation" for numbers of the relevant size? $\endgroup$
    – Igor Rivin
    Commented Sep 25, 2016 at 12:11
  • $\begingroup$ The degree of the discriminant is $n(n-1)$ so the typical size is $B^{n(n-1)}$ and we have only $2$ options modulo $4$ so the 2-valuation is $\geq k$ in about $B^{n(n-1)}/2^{k}$. Does this make sense? $\endgroup$
    – Lior Bary-Soroker
    Commented Sep 26, 2016 at 8:00
  • $\begingroup$ I made a mistake; the $n(n-1)$ is the degree in the roots, it should be replaced by $2n-2$, the degree in the coefficients $\endgroup$
    – Lior Bary-Soroker
    Commented Sep 26, 2016 at 10:02

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