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Clearly $ax + b \equiv 0 \pmod{p}$ is solvable for almost all primes $p$. In fact, if $p$ does not divide $a$, then $x \equiv -ba^{-1}$ is a solution. So for any linear polynomial $f(x)$ we get that for $100\%$ of the primes the equation $f(x) \equiv 0 \pmod{p}$ is solvable. With the use of quadratic reciprocity one can often relatively easy establish that for a given irreducible polynomial $f$ of degree two, for half of all primes the equation $f(x) \equiv 0 \pmod{p}$ has a solution, and I believe that this is true in general (?).

However, I now have the following polynomial of degree $d$:

$$f_d(x) = \sum_{i=0}^{d} \prod_{j = 0, j\neq i}^{d} (x+j)$$

And I would like to know the proportion of primes for which $f_d$ has a root. Is this something that can be calculated?

For some examples, with $d = 1, \ldots, 5$ we get:

$f_1(x) = 2x + 1$

$f_2(x) = 3x^2 + 6x + 2$

$f_3(x) = 4x^3 + 18x^2 + 22x + 6$

$f_4(x) = 5x^4 + 40x^3 + 105x^2 + 100x + 24$

$f_5(x) = 6x^5 + 75x^4 + 340x^3 + 675x^2 + 548x + 120$

For $f_1$ we have that $f_1(x) \equiv 0 \pmod{p}$ is solvable for all odd primes and it's a nice exercise to show that $f_2(x) \equiv 0 \pmod{p}$ is solvable whenever $p \equiv \pm 1 \pmod{12}$. Already for $f_3$ I have no idea what to do, let alone in general. From some searching around on the internet, it looks like Chebotarevs density theorem might be useful, but as of now I don't even understand its statement. Any help would be much appreciated!

Edit: as Will Jagy pointed out in the comments, for odd $d$ it looks like $f_d(x)$ is divisible by $2x + d$ or, alternatively, if $d = 2m+1$, then $f_d(x-m)$ is divisible by $2x+1$. Perhaps interestingly and at least for odd $d$ smaller than $10$, we have that the constant and linear terms of $f_d(x)/(2x + d)$ are equal to those of $f_{d-1}(x)$.

For even $d$ we can write $f_{d}(x-d/2) = \displaystyle \sum_{i=-d/2}^{d/2} \prod_{j = -d/2, j\neq i}^{d/2} (x+j)$ and then it is not hard to see that for $i_1 = -i_2$ the odd powers in the products for $i = i_1$ and $i = i_2$ cancel each other out, so $f_{d}(x+d/2)$ is the sum of even powers of $x$. For example:

$f_2(x) = 3(x+1)^2 - 1$

$f_4(x) = 5(x+2)^4 - 15(x+2)^2 + 4$

$f_6(x) = 7(x+3)^6 - 70(x+3)^4 + 147(x+3)^2 - 36$

$f_8(x) = 9(x+4)^8 - 210(x+4)^6 + 1365(x+4)^4 - 2460(x+4)^2 + 576$

$f_{10}(x) = 11(x+5)^{10} - 495(x+5)^8 + 7161(x+5)^6 - 38225(x+5)^4 + 63228(x+5)^2 - 14400$

Here one can note that these coefficients all seem divisible by small primes.

With the help of a computer I have checked that $f_d(x)$ is irreducible for even $d \le 200$ and for odd $d < 200$, $f_d(x)$ can be written as $(2x+d)$ times an irreducible.

Woett
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  • see theorem of Frobenius page 11 http://www.math.leidenuniv.nl/~hwl/papers/cheb.pdf – Will Jagy Sep 18 '18 at 23:11
  • so far, when $n$ is odd, the polynomial $f_n(x)$ has $2x+n$ as a factor – Will Jagy Sep 18 '18 at 23:31
  • and the quartic is a pure translation of a polynomial with only even exponents.... – Will Jagy Sep 18 '18 at 23:48
  • With a computer I just checked that for odd $n < 200$, $f_n(x)$ is equal to $2x + n$ times an irreducible polynomial. – Woett Sep 18 '18 at 23:54
  • And for $n = 6$ and $n = 8$ one can also write $f(x) = g(x+n/2)$ where $g$ has only even exponents. – Woett Sep 19 '18 at 00:20
  • Good. How about if you type in those translated versions for $n=4,6,8.$ There is, in some cases a favorable outcome to this with regard to primes represented by certain binary quadratic forms, or $\pm p$ if the binary is indefinite. See, for example, $x^2 - 3 y^2$ and your $\pm 1 \pmod {12}.$ Worth investigating, even though it cannot be the whole story in high degree $n$ – Will Jagy Sep 19 '18 at 00:26
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    Thanks for the translated polynomials. i am trying something along the lines of https://math.stackexchange.com/questions/2887405/set-of-primes-p-which-x4-x3-2x2-2x-1-completely-factors-in-finite-field-o for the quartic. I did not get that one until I had a modest list of all the factoring decomposition types, as in the Frobenius density theorem. That one was positive binary forms; your problem has indefinite binary forms, annoying when trying to recognize – Will Jagy Sep 19 '18 at 15:56
  • Will, first of all, thanks a lot for your help! Secondly, in the question you link, you seem to do much more than what I would need here, namely just the density of primes. For example, PARI/GP tells me that the Galois group of $f_4$ equals $D_4$, the dihedral group with $8$ elements (I assume? PARI/GP uses the so-called GAP4 name and calls it "D(4)"). Doesn't that by Frobenius already suffice to prove that $f_4(x) \equiv 0 \pmod{p}$ is solvable for $3/8$ths of the primes? – Woett Sep 20 '18 at 11:30
  • Yes, if all you want is the density $3/8,$ you are in good shape. – Will Jagy Sep 20 '18 at 17:37
  • Well, I want lots of things! Ideally for all $d$ I would like to know the density of primes $p$ for which $f_d(x) \equiv 0 \pmod{p}$ has a solution. Slightly less ambitiously is to find the density for small values of $d$ and possibly some upper and/or lower bounds for general $d$. Right now the following have my priority:

    1. Confirmation that the Galois group of $f_4(x)$ equals $D_4$ and that this implies density $3/8$ for $d = 4$.

    2. Finding the density for $d = 6,8,10$.

    3. (Mainly for theoretical and esthetic reasons) Proving that $2x + d$ divides $f_d(x)$ for all odd $d$l.

     – Woett Sep 20 '18 at 17:59
  • I think I get it now, and I might eventually (partially) answer my own question. Since $f_d(x)$ is a translate of a polynomial with even exponents, their Galois groups $G_d$ must coincide. For polynomials with even exponents, if $x$ is a root, so is $-x$. Therefore, if $\sigma \in G$ is such that $\sigma(x) = y$, then $\sigma(-x)$ must be equal to $-y$. This implies $|G_d| \le 2 \cdot 4 \ldots \cdot d = 2^{d/2} \cdot (d/2)!$ If equality holds (which it does for $d = 4, 6, 10$) then the number of elements $G_d$ with a cycle of length $1$ shouldn't be too hard to find, and we have our density. – Woett Sep 21 '18 at 11:28

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