# On the Cubic Pell equation $x^3+dy^3+d^2z^3-3dxyz =1$ for $d = 23$ etc.

(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.)

• 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 $c.$ – Will Jagy Sep 21 '13 at 23:32
• 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 – Tito Piezas III Sep 21 '13 at 23:58
• 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. – Will Jagy Sep 22 '13 at 1:29
• 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. – O. S. Dawg Nov 9 '14 at 1:10

## 1 Answer

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.

• If I am not mistaken, a few more largest "smallest" z's can be found at d = 239, 1301, 1721, 11621 and 33213. – O. S. Dawg Mar 18 '15 at 3:36
• @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. – Tito Piezas III Mar 18 '15 at 5:43
• 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. – O. S. Dawg Mar 18 '15 at 13:19
• 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. – O. S. Dawg Oct 5 '15 at 2:49