Two years ago, I made a conjecture on stackexchange:

Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

I have found some solutions, such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Here is another simple solution: $$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)?

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    $\begingroup$ It's a K3 surface, so you cannot give a complete rational parametrization; but there are enough elliptic fibrations that once you've found a few solutions you can probably bounce them around to get infinitely many others (e.g. from a typical solution you get three others by fixing two of the variables and finding the other solution for the third; now choose another variable and repeat). $\endgroup$ – Noam D. Elkies Jun 5 '15 at 1:25
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    $\begingroup$ I'm voting to close this question as off-topic because it has been answered in a comment. $\endgroup$ – Alex Degtyarev Jun 5 '15 at 7:33
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    $\begingroup$ @Alex, I don't understand. Why is an answer in a comment a reason for closure as off-topic? $\endgroup$ – Gerry Myerson Jun 5 '15 at 9:52
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    $\begingroup$ @Alex, any answer that starts with "It's a K3 surface" is research level (and not at all obvious) to me. $\endgroup$ – Gerry Myerson Jun 5 '15 at 10:06
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    $\begingroup$ @AlexDegtyarev Noam's comment didn't answer the question of how to find/describe all rational solutions, although it could have been given as an answer (rather than a comment) for how to find infinitely many, and possibly how to find a Zariski dense set of solutions. But in any case, characterizing rational points on K3 surfaces certainly qualifies as a research-level problem, and this seems like a nice example due to the symmetry. $\endgroup$ – Joe Silverman Jun 5 '15 at 17:10

Call (the projective completion of) your surface $S$. It admits three double covers of $\mathbb P^2$, namely $$\pi_1(a,b,c)=(a,b), \quad\pi_2(a,b,c)=(a,c),\quad\pi_3(a,b,c)=(b,c).$$ Each double cover induces an involution, so we get three involutions $$\sigma_1,\sigma_2,\sigma_3:S\to S.$$ These involutations don't commute, and if you form $f=\sigma_1\circ\sigma_2$ and $g=\sigma_1\circ\sigma_3$, then $f$ and $g$ generate a subgroup $G$ of $\text{Aut}(S)$ that is a free group on two generators. And presumably for most starting points $P\in S(\mathbb{Q})$, repeated application of $f$ and $g$ will give you a tree of rational solutions. (You might compare this with the generation of all positive integer solutions to the Hurwitz equation $x^2+y^2+z^2=3xyz$ starting from $(1,1,1)$.) Here are some references to papers that have studied rational points on K3 surfaces admitting 3 involutions:

  • Baragar, A. Rational points on K3 surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. Math. Ann. 305 (1996), no. 3, 541–558.

  • Wang, L., Rational Points and Canonical Heights on K3-surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$, Contemporary Math. 186 (1995), 273 – 289

  • $\begingroup$ Nice answer!,maybe this problem exist simple methods? $\endgroup$ – math110 Jun 6 '15 at 1:28

The original proposer asks for "simple methods". Simplicity, like beauty, is in the eye of the beholder. I am sure that Noam Elkies and Joe Silverman feel their answers are extremely simple. The following discussion is, in my humble opinion, simpler.

We can express the underlying equation as a quadratic in $a$, \begin{equation*} a^2+\frac{8bc}{(b^2-1)(c^2-1)}a-1 \end{equation*} with the obvious condition that $|b| \ne 1$ and $|c| \ne 1$.

For $a$ to be rational, the discriminant must be a rational square, so there exists $D \in \mathbb{Q}$ such that \begin{equation*} D^2=(c^2-1)^2b^4-2(c^4-10c^2+1)b^2+(c^2-1)^2 \end{equation*}

This quartic has an obvious rational point when $b=0$, and so is birationally equivalent to an elliptic curve. We find the curve \begin{equation*} v^2=u(u+(c^2-1)^2)(u+4c^2) \end{equation*} with the reverse transformation \begin{equation*} b=\frac{v}{(c^2-1)(u+4c^2)} \end{equation*}

The elliptic curve has $3$ points of order $2$, which give $b=0$ or $b$ undefined. There are also $4$ points of order $4$ at \begin{equation*} u=2c(c^2-1) \hspace{1cm} v= \pm 2c(c+1)(c-1)(c^2+2c-1) \end{equation*} and \begin{equation*} u=-2c(c^2-1) \hspace{1cm} v= \pm 2c(c+1)(c-1)(c^2-2c-1) \end{equation*} all of which give $|b|=1$.

Thus, to get a non-trivial solution we need the elliptic curve to have rank at least $1$. Numerical investigations suggest that the rank is often $0$, so solutions do not exist for all $c$.

We can derive parametric solutions by finding points of the curve, subject to certain conditions.

For example, $u=c^2-1$ would give a point if $5c^2-1=\Box$. We can parametrize this quadric using the solution when $c=1$, to give \begin{equation*} a=\frac{(p-2)(p-5)(3p-5)}{p(p-1)(p-3)(2p-5)} \hspace{1cm} b=\frac{p^2-4p+5}{2(p^2-5p+5)} \hspace{1cm} c=\frac{p^2-4p+5}{p^2-5} \end{equation*} which gives strictly positive solutions when $p > 5$.

Another simple point to consider could be $u=2c^2(c-1)(c+3)$ which gives a rational point when $(c+3)(3c+1)=\Box$.

  • $\begingroup$ It's nice solution!+1 $\endgroup$ – math110 Jun 11 '15 at 10:18
  • $\begingroup$ "I am sure that Noam Elkies and Joe Silverman feel their answers are extremely simple. The following discussion is, in my humble opinion, simpler." Hi Allan, maybe that's a bit snarky? In any case, your solution and my solution are complementary. You've explained how to transform the equation into an explicit elliptic surface, which is great. I explained a very general procedure for using one solution to find lots of others. It is also very simple, since it comes down to the fact that if you know one solution of a quadratic equation, it's easy to write down the other one! So using your ... $\endgroup$ – Joe Silverman Jun 11 '15 at 13:59
  • $\begingroup$ ... method to find lots of solutions involves the group law on elliptic curves, while the solution I described, albeit without working out the explicit equations, involves solving quadratic polynomials. So despite my admiration for the theory of elliptic curves, I think it would actually be easier to explain the quadratic equation method to, say, a high school student. OTOH, the rational curves you've found are completely explicit, which is even better. $\endgroup$ – Joe Silverman Jun 11 '15 at 14:05
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    $\begingroup$ I can assure you that no offence was meant, if I guess correctly what "snarky" means - never heard the word before! I was just trying to point out that, to experts such as yourself and Noam Elkies, K3 Surfaces are simple objects, but not perhaps to others. $\endgroup$ – Allan MacLeod Jun 11 '15 at 15:48

@Allan methods it's nice! here is my answer: since $$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$ so let $$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{z}=\dfrac{1-b^2}{2b},\;\dfrac{z}{x}=\dfrac{1-c^2}{2c}$$ and solving for $a,b,c$, $$a = \frac{-x+\sqrt{x^2+y^2}}{y},\;\;b = \frac{-y+\sqrt{y^2+z^2}}{z},\;\; c = \frac{-z+\sqrt{x^2+z^2}}{x}$$ it is easy to see $x^2+y^2,y^2+z^2,z^2+x^2$ must be square. so we use Euler bricks solution $$x=u|4v^2-w^2|,y=v|4u^2-w^2|,z=4uvw$$ then it is not hard to find to give solution



I'm late for this party, but using math110's method employing Euler bricks, couldn't resist giving some simple rational solutions to,

$$(1-a^2)(1-b^2)(1-c^2) = 8abc$$

Solution 1:

$$a,\,b,\,c = \frac{-(x-z)(2x+z)}{(2x-z)y},\;\frac{z}{2x},\;\frac{-2y+z}{2y+z}\tag1$$

where $x^2+y^2=z^2.$

Solution 2:

$$a,\,b,\,c = \frac{2z^2}{xy},\;\frac{x-z}{x+z},\;-\frac{y+z}{y-z}\tag2$$

where $x^2+y^2=5z^2$, and which may also be solved as a Pell equation.


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