Cube roots of five

Question

This is not really homework.

I might be able to do this myself in time, from the methods in Ireland and Rosen.

Note that every number has exactly one cube root $\pmod q$ for any prime $q \equiv 2 \pmod 3.$

For primes larger than $3$ with $p \equiv 1 \pmod 3:$ Gauss: $2$ is a cube if and only if we can write $p = u^2 + 27 v^2,$ in which case there are three distinct cube roots of $2.$ Jacobi: $3$ is a cube if and only if we can write $p = u^2 + uv + 61 v^2,$ in which case there are three distinct cube roots of $3.$

I would like a proof for this, already checked on computer; for primes larger than $3$ with $p \equiv 1 \pmod 3:$ $5$ is a cube $\pmod p$ if and only if we can write $p = u^2 + uv + 169 v^2$ OR $p = 13u^2 + uv + 13 v^2,$ in which case there are three distinct cube roots of $5.$

EEDDIITT, Tuesday: I would also like a proof that for primes larger than $10$ with $p \equiv 1 \pmod 3:$ $7$ is a cube $\pmod p$ if and only if we can write $p = u^2 + uv + 331 v^2$ OR $p = 19u^2 + 11uv + 19 v^2,$ in which case there are three distinct cube roots of $7.$

I think it very likely that these can be done entirely using the methods in chapter 9 of Ireland and Rosen, chapter title "Cubic and Biquadratic Reciprocity." I did use the tag class-field-theory but I think that unnecessary. Cubic seems to be just pages 108-119, then some of the exercises 134-137. I will see what I can do.

NOTE: the full class group for discriminant $-675 $ has two genera: the principal genus has forms (classes) $$ \langle1,1,169 \rangle; \; \; \langle9,3,19 \rangle; \; \; \langle9,-3,19 \rangle. $$ Note that these forms always give residues $\pmod 5$ when they are not divisible by $5.$

The other genus has forms (classes) $$ \langle13,1,13 \rangle; \; \; \langle7,5,25 \rangle; \; \; \langle7,-5,25 \rangle. $$ Note that these forms always give non-residues $\pmod 5$ when they are not divisible by $5.$

Motivation: i am interested in finding out the numbers that have an integral representation as $9 x^2 + 3 x y + 19 y^2 - z^3.$ There is an interesting wrinkle here, there are more negative numbers, the absolute values of those numbers not represented (apparently) give a superset of the postive ones. i think it is caused by having more than one genus. I already believe that all numbers not represented are of the forms $5 A^3$ and $3125 B^3,$ but there are more negative $A$ than positive $A,$ it seems, probably similar for $B.$

LATE Sunday: I figured out the apparent lack of symmetry. I did not let the computer go high enough, that's all. All of them are refuted by three formulas together with Gauss composition, $$ 6^3 - 5 = 211 = x^2 + x y + 169 y^2, \; x=6,y=1 $$ $$ 12^3 - 5 = 1723 = 13 x^2 + x y + 13 y^2, \; x=11,y=3 $$ $$ 23^3 - 40 = 12127 = 13 x^2 + x y + 13 y^2, \; x=30,y=-7 $$

Compare What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$

Answer 1

Yes, these can be done with cubic reciprocity. This is following Cox, Primes of the form $x^2+ny^2$ $\S 4.A$.

In the sequel let $a,b,c,p,r,s,u,v$ denote rational integers.

Let $\omega=e^{2\pi i/3}$, and $\Bbb Z[\omega]$ is a Euclidean domain with norm $N(a+b\omega)=a^2-ab+b^2$.

Let $\pi,\alpha,\beta,\sigma$ denote elements of $\Bbb Z[\omega]$.

If $\pi$ is prime in $\Bbb Z[\omega]$ then $\pi$ is called primary if $\pi=a+3b\omega$, and every prime has a primary associate.

If $\pi$ is prime and $\pi \not \mid 3\alpha\beta$ then cubic reciprocity says $$ \left(\frac\alpha\pi\right)_3 \equiv \alpha^{(N(\pi)-1)/3} \pmod \pi \\ \left(\frac{\alpha\beta}\pi\right)_3 = \left(\frac\alpha\pi\right)_3 \left(\frac\beta\pi\right)_3 \\ \left(\frac\alpha\pi\right)_3 = \left(\frac\beta\pi\right)_3 \quad \mathrm{whenever}~\alpha\equiv\beta\pmod\pi $$ and if $\pi,\alpha$ are primary primes then $$ \left(\frac\alpha\pi\right)_3 = \left(\frac\pi\alpha\right)_3 $$

If $p\equiv 1\pmod 3$ is a rational prime then we can write $p=\pi \overline\pi$ with $\pi$ and $\overline\pi$ primary primes. Since $5$ is also a primary prime in $\Bbb Z[\omega]$ with $N(5)=25$, write $\pi=a+3b\omega$ then $$ p = a^2-3ab+9b^2\\ \left(\frac 5\pi\right)_3 =\left(\frac\pi 5 \right)_3 \equiv (a+3b\omega)^8 \pmod 5\\ \left(\frac 5\pi\right)_3 = 1 \iff (a+3b\omega)^4\equiv \pm 1 \pmod 5 \\ \iff a^4-3a^3b\omega-a^2b^2\omega^2-2ab^3+b^4\omega \equiv \pm 1 \pmod 5 \\ a^4+a^2b^2-2ab^3+(b^4+a^2b^2-3a^3b)\omega \equiv \pm 1 \pmod 5 \\ \iff 5 \mid a^4+a^2b^2-2ab^3\pm1 \quad \mathrm{and} \quad 5 \mid b^4+a^2b^2+2a^3b \\ b\equiv 0 \quad \mathrm{or} \quad a\equiv -b \pmod 5 $$ which we can get by enumerating cases. In the first case write $b=5c$ then $$ p = a^2-15ac+225c^2 = u^2+uv+169v^2 $$ with $u=a-7c,v=-c$. In the second case write $b=5c-a$ then $$ p = 13a^2-105ac+225c^2 = 13u^2+uv+13v^2 $$ with $u=4c-a,v=c$. It's immediate that the conditions are the same for $(5/\overline\pi)_3=1$ and hence for $5$ to be a cube mod $p$.

Since $7=\sigma \overline\sigma = (1+3\omega)(-2-3\omega)$ is not prime in $\Bbb Z[\omega]$ we proceed a bit differently. Taking $7<p=\pi\overline\pi$ as above $$ \pi=a+3b\omega \equiv a+3b\omega-b(1+3\omega) = a-b \pmod \sigma \\ \pi \equiv a+3b\omega+b(-2-3\omega) = a-2b \pmod {\overline\sigma} \\ \left(\frac \sigma\pi\right)_3 = \left(\frac\pi\sigma\right)_3 \equiv (a+3b\omega)^2 \pmod \sigma $$ Since $\sigma\mid 7$, for any $\alpha\in\Bbb Z[\omega]$ with $\sigma \not\mid \alpha$ we can write $\alpha \equiv r \pmod \sigma$ for some $r\in\{\pm 1,\pm 2,\pm 4\}$. It is straightforward to find $$ \left(\frac{\pm 1}\sigma\right)_3 = 1, \quad \left(\frac{\pm 2}\sigma\right)_3 = \omega^2, \quad \left(\frac{\pm 4}\sigma\right)_3 = \omega, \quad \\ \left(\frac{\pm 1}{\overline\sigma}\right)_3 = 1, \quad \left(\frac{\pm 2}{\overline\sigma}\right)_3 = \omega, \quad \left(\frac{\pm 4}{\overline\sigma}\right)_3 = \omega^2, \quad $$ So $$ \left(\frac 7\pi\right)_3 = \left(\frac\sigma\pi\right)_3\left(\frac{\overline\sigma}\pi\right)_3 = \left(\frac\pi\sigma\right)_3\left(\frac\pi{\overline\sigma}\right)_3 = \left(\frac{a-b}\sigma\right)_3\left(\frac{a-2b}{\overline\sigma}\right)_3 = 1 \\ \iff a-b\equiv \pm (a-2b) \pmod 7 \\ b\equiv 0 \quad \mathrm{or} \quad b\equiv 3a \pmod 7 $$

In the first case write $b=7c$ then $$ p = a^2-21ac+441c^2 = u^2+uv+331v^2 $$ where $u=a-10c,v=-c$. In the second case write $b=7c+3a$ then $$ p = 73a^2-357ac+441c^2 = 19u^2+11uv+19v^2 $$ where $u=a-2c,v=5c-2a$. Again it follows that one of these is necessary and sufficient for $7$ to be a cube mod $p$.