# Why are some solutions of these diophantine equations off the usual patterns?

This is inspired by a recent question about complete multipartite integral graphs. I am wondering if more can be said about tripartite integral graphs with block sizes $a<b<c$. It is easy to see that their three nonzero eigenvalues are the same as those of $$\begin{pmatrix} 0&b&c\\ a&0&c\\ a&b&0 \end{pmatrix}$$ because the corresponding eigenvectors must have the same entry inside each block. From Cardano's formula we get as a necessary condition for integer eigenvalues of this $3\times3$ matrix that $$\Delta=(ab+ac+bc)^3-27(abc)^2$$must be a square.
(Or, in terms of their reciprocals $u,v,w$, which can be wlog required to be integers, $$\Delta'=\frac{(u+v+w)^3}{uvw}-27$$ must be a square.)
Note that this condition is not sufficient; e.g. for $c<550$ it is fulfilled by $86$ coprime triples, but only for $71$ of them the eigenvalues are integers.

Here are the triples for $c<550$.

          a,b      c             λ1    λ2   λ3
5     8    12           -10    -6   16
*   5    12    77           -33    -7   40
13     4    48           -26    -6   32
13     7    45           -26    -9   35
13    24    28           -26   -16   42
17     3    65           -34    -5   39
17    12    56           -34   -14   48
17    33    35           -34   -21   55
25     9    91           -50   -13   63
25    12    88           -50   -16   66
29    36    80           -58   -32   90
29    39    77           -58   -33   91
37    15   133           -74   -21   95
37    16   132           -74   -22   96
37    63    85           -74   -45  119
41    11   153           -82   -17   99
41    24   140           -82   -30  112
41    60   104           -82   -48  130
*  45   136   396          -240   -66  306
*  49   220   441          -315   -77  392
53     5   207          -106    -9  115
53    44   168          -106   -48  154
53    95   117          -106   -65  171
61    13   231          -122   -21  143
61    40   204          -122   -48  170
61    84   160          -122   -70  192
65     8   252          -130   -14  144
65    51   209          -130   -57  187
65    99   161          -130   -77  207
73    84   208          -146   -78  224
73   105   187          -146   -85  231
85    21   319          -170   -33  203
85   112   228          -170   -96  266
89    20   336          -178   -32  210
89    57   299          -178   -69  247
89   176   180          -178  -110  288
97   115   273          -194  -105  299
97   136   252          -194  -112  306
101    63   341          -202   -77  279
101   143   261          -202  -117  319
101    24   380          -202   -38  240
109    96   340          -218  -102  320
109   189   247          -218  -133  351
109     7   429          -218   -13  231
113   144   308          -226  -126  352
113    84   368          -226   -96  322
113    17   435          -226   -29  255
125   221   279          -250  -153  403
*  125   357   500          -425  -175  600
137   203   345          -274  -161  435
137   152   396          -274  -144  418
145   112   468          -290  -126  416
145    99   481          -290  -117  407
149   260   336          -298  -182  480
149    69   527          -298   -93  391
149    56   540          -298   -80  378
157   276   352          -314  -192  506
157   135   493          -314  -145  459
169   220   456          -338  -190  528
169   217   459          -338  -189  527
173   285   407          -346  -209  555
181   220   504          -362  -198  560
185   308   432          -370  -224  594
193   364   408          -386  -238  624
197   255   533          -394  -221  615
* * 200    44   525          -330   -70  400
205   391   429          -410  -253  663
*  225   200   252          -240  -210  450
233   464   468          -466  -288  754
241   451   513          -482  -297  779


Note that most often, $\lambda_1=-2a$ or $\lambda_1=-2b$ (BTW the table is sorted by this $a$ or $b$ in the first column, sometimes swapping $a$ and $b$). If we denote the entries of the first column by $a$, all these values have only divisors of the form $4k+1$, moreover each time $b+c=4a$. Further, $\lambda_2$ (and also $\lambda_3$, because $\lambda_1$ is even and the trace is $0$) has the same parity as $b$. I suppose all this can be explained easily by somebody who is knowledgeable about the properties of the related elliptical curve.
But the really intriguing thing are the exceptions where $-\lambda_1$ is not the double of $a$ or $b$. They are marked by $*$.

• What about these exceptional triples? Are there infinitely many of them?

• For the regular triples, does each prime $p=4k+1$ occur at least three times as one of $a$ or $b$?
(So far, $13$ and $17$ even occur four times)

• Are there infinite families of co-prime triples which can be explicitely given (e.g. by a recursion)?

I have also done a similar exhaustive search for 4-partite graphs with block sizes $a<b<c<d$. For $d<80$, there are none, but there are some, e.g. $K_{8,15,32,40\ \ }$ or $K_{3,35,48,77\ \ }$, which have two integer nonzero eigenvalues.

• Do such 4-partite integral graphs exist?
• Along $b + c =4a$, $\Delta$ is a square times $b^2 + 18 bc + c^2$, so there is a solvable conic (isomorphic to $b^2 + 18bc + c^2 = z^2$) above the line $b + c = 4a$ in the surface $\Delta(a,b,c) = z^2$. This conic may well account for the majority of the rational points on your surface. – Michael Stoll Dec 7 '15 at 12:11
• If $p$ is an odd prime divisor of $a$ and $b+c = 4a$, then $c \equiv -b \bmod p$, hence $p \mid (4b)^2 + z^2$, so $p \equiv 1 \bmod 4$. – Michael Stoll Dec 7 '15 at 12:14
• @MichaelStoll You are right, so in fact the "regular" solutions are easy, as they can be cast as solutions of $(b+c)^2+16bc=z^2$. But isn't it puzzling that there are those "sporadic" solutions beside them? – Wolfgang Dec 7 '15 at 12:40

The question is about rational points on the surface $S$ given by $$\Delta(a,b,c) := (ab+bc+ca)^3 - 27(abc)^2 = z^2$$ in the weighted projective space ${\mathbb P}_{1,1,1,3}$. Along the line $b+c = 4a$ the left hand side factors as a square times $b^2 + 18bc + c^2$, and the equation effectively reduces to $b^2 + 18bc + c^2 = z^2$, which is a solvable conic and has lots of rational points that can be parameterized. Also, for these points, any odd prime divisor of $a$ satisfies $p \mid (4b)^2 + z^2$, which implies (for coprime $a,b,c$) that $p \equiv 1 \bmod 4$.

There are other curves on $S$. For example, considering lines through the point $(1:1:1)$ (which is a simple double point of (the rational curve) $\Delta = 0$), we obtain curves of the form $z^2 = q(x,y)$, where $q$ is a quartic. If we take the line $4a = b + 3c$, this results in a curve birational to $$y^2 = 64 x^4 + 480 x^3 + 912 x^2 + 640 x + 117,$$ which is a curve of genus 1 with some rational points. Doing the math, we find that this curve is isomorphic to the elliptic curve $y^2 + y = x^3 - 1$ (243a1 in the Cremona database) of Mordell-Weil rank 1. This implies that there are infinitely many rational points on this curve. The smallest points with strictly positive coordinates coming from this curve have (after re-ordering) $$(a,b,c) = (2,3,6), \quad (6,57,74), \quad (2068,7657,9520).$$

More generally, considering the pencil of lines through $(1:1:1)$ gives us a fibration of $S$ into genus 1 curves that admits a horiziontal curve with a degree 2 map to the base; this curve is the conic from above. So after a base-change of degree 2, we actually have a fibration into elliptic curves, and one would expect at least half of them to have positive rank, resulting in infinitely many genus 1 curves on $S$ with infinitely many rational points on each. For example, your first starred solution $(5, 12, 77)$ is on a curve isomorphic to $y^2 = x^3 - 41659288807500$, which has even rank 2.

• Thank you! For 4a=b+3c, the 3rd point (2068,7657,9520) even yields integer eigenvalues. :) – Wolfgang Dec 7 '15 at 14:08