4
$\begingroup$

(This post was motivated by an old one.) For Pell equations,

$$x^2-dy^2 = 1\tag{1}$$

and $d<100$, the largest fundamental solution is for $d = 61$ (which happens to be the 6th power of a fundamental unit),

$$x+y\sqrt{61} = 1766319049+226153980\sqrt{61} =\left(\frac{39+5\sqrt{61}}{2}\right)^6$$

In general, for prime $d = 8n+5$ and odd fundamental solution to $u^2-dv^2=-4$, then initial solution to $(1)$ will be a 6th power and thus huge. For the cubic analogue,

$$x^3+dy^3+d^2z^3-3dxyz = 1\tag{2}$$

just like the Pell, from an initial solution, an infinite more can be found. But how do we characterize "tricky" $d$ such that the smallest positive $x,y,z$ of $(2)$ are in fact relatively large. (From limited data, I think prime $d = 12n-1$ is one subset).

Question: Anybody knows how to find $d=23,47,59,71$ of $(2)$?

(P.S.1 It would suffice to find "small" signed $x,y,z$ since one can derive the positive ones from those. For example, for $d=11$ and $x,y,z = 1,4,-2$, one can derive $x,y,z = 89, 40, 18$.

(P.S.2 Useful details can be found in Springer's The Cubic Analogue of Pell's Equation.)

$\endgroup$
  • $\begingroup$ just a little culture, for your $c=2$ math.stackexchange.com/questions/329936/… while for $c=3$ the primes represented would be $p = x^2 + x y + 61 y^2$ whenever $p \equiv 1 \pmod 3.$ I don't know anything so pretty for larger $c.$ $\endgroup$ – Will Jagy Sep 21 '13 at 23:32
  • $\begingroup$ Ah, your linked post asks when $F(x,y,z)= x^3+cy^3+c^2z^3-3cxyz = p$, for some prime $p$ for $c = 2$, and $c=3$. My interest started when considering if there is a polynomial identity for $v_1^3+v_2^3+v_3^3 = Nv_4^3$ where $v_4 =F(x,y,z)=1$. See the context in this MO post, Integers as sums of three integer cubes $\endgroup$ – Tito Piezas III Sep 21 '13 at 23:58
  • $\begingroup$ Right, I enjoy finding what numbers are integrally represented by some polynomial in a few variables, especially when the product of two represented numbers is also represented, as is the case with your polynomials above. $\endgroup$ – Will Jagy Sep 22 '13 at 1:29
  • $\begingroup$ I computed solutions for the first 1000 primes. Those primes of the form 12k+1 averaged 378.1 digits in length, 12k+5 ~ 1173.5, 12k+7 ~ 328.8 and primes of the form 12k+11 averaged 1070.1 digits. My solutions are not guaranteed to be minimal. $\endgroup$ – O. S. Dawg Nov 9 '14 at 1:10
1
$\begingroup$

The question was to solve,

$$x^3+dy^3+d^2z^3-3dxyz = 1\tag{1}$$

for $d = 23, 47, 59, 71$. (Note: The equation $(1)$ can be solved in the integers for all non-cube integer $d$.) Seiji Tomita has done more and created a table of fundamental solutions for non-cube $d<100$. The largest "smallest" $x,y,z$ found so far is for $d = 69 = 3\cdot23$,

$$x,y,z ={13753611475894008059401,\;-5630668308465438120720,\; 555253697459615284770}$$

From this, one can derive the "smallest" positive solution to $(1)$ for $d = 69$ (and from which an infinite more can then be found),

$$x,y,z = 404886837053487091694212951195653956127452401,\; 98715184393700556938337454013404500951638820,\; 24067681974543893805323831567684099602695630$$

Analogous to Pell equations, the $x/y, y/z$ are close to $69^{1/3}$ (within $10^{-68},10^{-67}$, respectively). The ratios of larger positive $x,y,z$ will get ever closer and closer to $d^{1/3}$.

P.S. Pell equations appear in a lot of contexts, and can solve other Diophantine equations, like $x_1^3+x_2^3+x_3^3=1$. Right now $(1)$ seems to be only a mathematical curiosity, but maybe someday, someone, somewhere can use it to prove that such-and-such equation has an infinite number of integer solutions.

$\endgroup$
  • $\begingroup$ If I am not mistaken, a few more largest "smallest" z's can be found at d = 239, 1301, 1721, 11621 and 33213. $\endgroup$ – O. S. Dawg Mar 18 '15 at 3:36
  • $\begingroup$ @O.S.Dawg: By coincidence, I've been meaning to ask a question about the similar $x^3+y^3 = Nz^3$. You may be interested in this post. $\endgroup$ – Tito Piezas III Mar 18 '15 at 5:43
  • $\begingroup$ Yes, that post is interesting. For the cubic pell I went looking for d with solutions having more than d digits. Best so far: d=32213 with 31041 digits in the apparent smallest positive solution. Some suspects I will try: 39521, 92009, 143879 and 343433. $\endgroup$ – O. S. Dawg Mar 18 '15 at 13:19
  • $\begingroup$ Had a go at computing solutions for d = 39521 and d = 92009 this weekend. For the first I get a 40027 digit solution and the second I found a 94980 digit solution. Not a big surprise. Working on the latter two. $\endgroup$ – O. S. Dawg Oct 5 '15 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.