4

It seems the solution of,

$$u^2+dv^2 = w^3\tag1$$

involves the class number $h(d)$. Assume $\gcd(u,v)=1$.

Q: For which $\color{red}{prime}\; d$ is the complete solution of $(1)$ in the integers given by,

$$(p^3 - 3 d p q^2)^2 + d(3 p^2 q - d q^3)^2 = (p^2+dq^2)^3\tag2$$

Equating terms between $(1)$ and $(2)$ and eliminating $q$, then the question is equivalent to solving,

$$u + 3 w p - 4 p^3 = 0\tag3$$

Using Mathematica to test various $d$, it seems it depends $\text{mod}\, 8$ and on the class number $h(d)$,

  1. If $d=8n+1,8n+5$ with $h(4d)\neq3m$, then $(2)$ is complete with integer $p,q$. Ex. $d=5$.
  2. If $d=8n+1,8n+5$ with $h(4d)=3m$, then $(2)$ is not complete. Ex. $d=29$.
  3. If $d=8n+3$, then $(2)$ is complete with integer and half-integer $p,q$. Ex. $d=11,163$.
  4. If $d=8n+7$, then $(2)$ is not complete. Ex. $d=7,23$.

Q: Is $\text{#}1,2,3,4$ true?

Tito Piezas III
  • 60,745
  • I see you deleted $d=11.$ I'm not sure yet what you meant by half integers, but the thing does work out for $u^2 + uv + 3 v^2 = z^3.$ – Will Jagy Mar 28 '16 at 00:12
  • 1
    The argument is the 1923 paper of Dickson that Kieran asked me about in his recent question. That material is repeated in his 1929 book, Introduction to the Theory of Numbers, pages 91-96 mostly. I see how he has an exercise on page 96, solve $x^2 + y^2 = uvw$ which is your item above when $u=v=w$ – Will Jagy Mar 28 '16 at 00:21
  • @WillJagy: I only deleted some $d$ to unclutter the post. I'll restore $d=11$. By "half-integer", I mean that for any co-prime $u,v$, then one can *always* find $p,q = \frac{p'}{2}, \frac{q'}{2}$. – Tito Piezas III Mar 28 '16 at 00:43
  • @WillJagy: For example, given $4^2+11*1^2 = 3^3$, then the only solution is $p,q =\pm\frac{1}{2},\pm\frac{1}{2}$. – Tito Piezas III Mar 28 '16 at 00:52
  • Oh, alright. Actually, given any $w = x^2 + xy + 3 y^2,$ we can solve $u^2 + 11 v^2 = 4 w$ in integers. – Will Jagy Mar 28 '16 at 00:54
  • 1
    @WillJagy: Equivalently, $\big(\frac{u}{2})^2+11\big(\frac{v}{2})^2=x^2+xy+3y^2$. Hence the half-integers on the LHS. – Tito Piezas III Mar 28 '16 at 00:59
  • @WillJagy: I guess this also means $\big(\frac{u}{2})^2+163\big(\frac{v}{2})^2=x^2+xy+41y^2$ is solvable in the integers, right? – Tito Piezas III Mar 28 '16 at 01:02
  • Yes, the 41 version is good. – Will Jagy Mar 28 '16 at 01:04
  • this is not really surprising since elements of order $3$ in the class group correspond to solutions out of that parametrization. I guess when you take the wrong ring of integers, some weird stuff happens that either give trivial counterexamples or on the contrary make counterexamples have both even coordinates so that you don't consider them. – mercio Apr 07 '16 at 19:48

0 Answers0