Analysis of a quadratic diophantine equation

Question

Hi! This is my first post on Math Overflow. I have two equations: $a(3a-1) + b(3b-1) = c(3c-1)$ and $a(3a-1) - b(3b-1) = d(3d-1)$. I'm trying to find properties of $a$ and $b$ that lead to solutions, where $a, b, c, d \in \mathbb{N}$. I'm having trouble applying any of the techniques in my abstract algebra book, as they mostly only apply to linear Diophantine equations.

So far, I only really have managed to deduce the following things:

$2b(3b-1) + d(3d-1) = a(3a-1) +b(3b-1)$

$2b(3b-1) = c(3c-1) - d(3d-1)$

$2b(3b-1) = (c-d)(3(c+d)-1)$

Any ideas on where to go from here would be greatly appreciated. Thanks!

Answer 1

One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_1=6a-1$ etc. Then the variables $a_1$ etc are constrained to be congruent to $5$ modulo $6$.

Homogenizing these gives $$X^2+Y^2=Z^2+T^2$$ and $$X^2-Y^2=Z^2-T^2.$$ Searching for rational solutions of your equation is essentially looking for rational points on the intersection of these two quadrics in $\mathbf{P}^3$. In general the intersection of two quadrics in $\mathbf{P}^3$ is an elliptic curve, so it looks like your problem will boil down to something like finding the integer points on an elliptic curve.

Added There's a blunder in the above: I must thank Fedor for noticing that the second equation should be $$X^2-Y^2=W^2-T^2.$$ So the variety is the intersection of two quadrics in $\mathbf{P}^4$. Hartshorne mentions in passing that in general this construction gives a del Pezzo surface. Del Pezzo surfaces are rational so there should be a birational parametrizion (in terms of two affine parameters) of the rational solutions to the original pair of equations.

Answer 2

Completing the squares and setting $X = 6a-1$, $Y = 6b-1$, $Z = 6c-1$ and $W = 6d-1$ we find $$ X^2 + Y^2 = Z^2 + 1, \quad X^2 - Y^2 = W^2 - 1. $$

Adding these equations we get $$ 2X^2 = Z^2 + W^2, $$ which we can parametrize by $$ X = t^2 - 2tu + 2u^2, \quad W = t^2 - 4tu + 2u^2, \quad Z = t^2 - 2u^2. $$ This parametrization yields integral solutions as $t$ and $u$ run through ${\mathbb Z}$, although perhaps not all of them since these may have a common divisor.

Neglecting this problem for now we can plug this parametrization into $$ 2Y^2 = Z^2 - W^2 + 2 $$ and get $ Y^2 = 4ut^3 - 12u^2t^2 + 8u^3t + 1 $, that is, $$ (1) \qquad \qquad Y^2 = 4tu(t-u)(t-2u) + 1. $$

1. Brute Force A direct search for points on this surface yields several solutions; the solutions with $t < (2-\sqrt{2}\,)u$ give rise to values of $X$, $Y$, $-Z$, $W$ that are positive, and if $t \equiv 3 \bmod 6$ and $u \equiv \pm 1 \bmod 3$ these values are all $\equiv -1 \bmod 6$, hence yield positive integral solutions $(a,b,c,d)$ of our problem The two smallest out of more than a dozen found this way are $(t,u,a,b,c,d) = (9 , 85, 2167, 1020, 2395, 1912)$ and
$(t,u,a,b,c,d) = (51, 2506, 2051177, 415877, 2092912, 2008575)$.

This suggests that the diophantine problem has infinitely many solutions in positive integers. Proving this conjecture seems to be difficult, however.

2. Elliptic Surfaces

Since $Y$ must be odd, we can set $Y = 2y+1$ and get $$ y^2 + y = ut^3 - 3u^2t^2 + 2u^3t = tu(t-u)(t-2u). $$ This is an elliptic surface, i.e., an elliptic curve over the field ${\mathbb Q}(u)$. Obvious rational points are $(t,y) = (0,0), (0,-1), (u,0), (u,-1), (2u,0), (2u,-1)$.

For transforming this into Weierstrass form, multiply through by $u^2$ and set $yu = z$, $tu = x$, giving $$ z^2 + uz = x^3 - 3u^2x^2 + 2u^4x = x(x-u^2)(x-2u^2). $$ The six rational points we had found above now are $P_1 = (0,0)$, $-P_1 = (0,-u)$, $P_2 = (u^2,0)$, $-P_2 = (u^2,-u)$, $P_3 = (2u^2,0)$, $-P_3 = (2u^2,-u)$ Observe that $P_1 + P_2 + P_3 = 0$; the points $P_1$ and $P_2$ generate a subgroup of $E({\mathbb Q}(u))$ with rank $2$.

A simple calculation shows $$ 2(0,0) = (4u^6 + 3u^2, -8u^9 - 6u^5 - u), $$ $$ 2(0,-u) = (4u^6 + 3u^2, 8u^9 + 6u^5), $$ $$ 2(u^2,0) = (u^6 + u^2, u^9 - u), $$ $$ 2(u^2,-u) = (u^6 + u^2, -u^9), $$ $$ 2(2u^2,0) = (4u^6-u^2,-8u^9 + 6u^5 - u) $$ $$ 2(2u^2,-u) = (4u^6-u^2,8u^9-6u^5) $$ These points provide us with the following parametrized families of integral points on our surface: $$(X,Y,Z,W) = (u^5 + u, 2u^4 - 1, u^5 + 2u^3 - u, u^5 - 2u^3 - u) $$ $$(X,Y,Z,W) = (16u^5 + 16u^3 + 5u, 16u^4+12u^2+1, 16u^5 + 24u^3 + 7u, 16u^5 + 8u^3 - u)$$ $$(X,Y,Z,W) = (16u^5 - 16u^3 + 5u, 16u^4-12u^2+1, 16u^5-8u^3-u, 16u^5-24u^3+7u) $$ None of these give us solutions to our original equation, however.

3. The Fibonacci connection. Since $E$ has rank $2$ over the function field ${\mathbb Q}(u)$, the elliptic curves $E_u$ will have rank $\ge 2$ except for at most finitely many exceptions (the only one I noticed is $u = 1$). For some families of specializations, the rank may be higher. This is the case if we take $u = F_n$, the $n$-th Fibonacci number. In this family, we have a point independent from the $P_j$ listed above, which means they have at least rank $3$ (except for the finitely many exceptions mentioned before). The points (modulo typos) are $$ Q = (F_{2n-2} F_{2n}, F_{2n-2} F_{2n}F_{2n+1}) $$ if $u = F_{2n}$, and $$ Q = (F_{2n-1}F_{2n+1}, F_{2n} F_{2n+1}^2) $$ if $u = F_{2n+1}$. While this does not seem to help us, I thought I'd mention it anyway since no one expects the Spanish inqui^H Fibonacci numbers in this problem.

Answer 3

For this system one can find a general rational parametrization and then N&S conditions for integer solutions.

Adding the pair:

$x^2 + y^2 = z^2 + 1$

$x^2 - y^2 = t^2 - 1$

gives:

$2 x^2 = z^2 + t^2$

which has a general rational parametrization (GPR):

$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$

$(z - t)/2 = 2 v x / (v^2 + 1)$

Adding these gives an expression for z and plugging this back in the first of the original pair then gives:

$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)

which has GPR:

$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$

and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:

$D = u^2 - v (v^2 - 1)$

$D x = u (v^2 + 1)$

$D y = u^2 + v (v^2 - 1)$

$D z = u (v^2 + 2 v - 1)$

$D t = u (v^2 - 2 v - 1)$

Homogenizing these by taking $u, v = a/c, b/c$ with $(a, b, c) = 1$, we now investigate how to specialize this to integer solutions.

First, $y$ is an integer iff $a^2 c - b (b^2 - c^2)$ divides $2 a^2 c$. Equivalently, there is an integer $L$ such that:

$b L (b^2 - c^2) = (L - 2) a^2 c$

Then two cases arise, depending on the parity of L.

Case 1 L odd

We show that this is impossible (given the other constraints of the problem).

If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:

$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$

Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:

$2 a b z = a^2 (L - 2) + 2 b^2 L$

$2 a b t = a^2 (L - 2) - 2 b^2 L$

So letting $a, b = A e, B e$ with $(A, B) = 1$ implies the following, in which $2 L / A$ and $(L - 2) / B$ are integers:

$2 z = A (L - 2) / B + B (2 L / A)$

$2 t = A (L - 2) / B - B (2 L / A)$

For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.

If they are both odd then A and $2 L / A$ must be both odd, which is impossible.

If they are both even then A even implies B odd and thus $2 L / A$ even, and A odd implies $(L - 2) / B$ even. So in either case this implies L even, contrary to hypothesis.

So that leaves us with ..

Case 2 L even

Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :

$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$ [*]

which, as in Case 1, implies:

$2 z = A (m - 1) / B + B (2 m / A)$

$2 t = A (m - 1) / B - B (2 m / A)$

Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..

Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.

So they must be both even, which is the case iff $A \equiv m \mod(2)$ (provided that when $m$ is odd, $(m - 1) / B$ is even, in other words $B$ does not divide out the power of 2 dividing $m - 1$).

Furthermore from the form of $z$, $t$, as $f \pm g$, they have the same parity. So adding and subtracting the original pair implies that $x, y$ are integers iff $z, t$ are integers.

Note that the above isn't an explicit integer solution. All I have done is reduce the problem to the pair [*], to which I have a draft solution that needs checking. But if anyone else wishes to nip in first with a solution to these then obviously feel free!

Answer 4

As Robin and Fedor observed the variety in question is a quartic Del Pezzo surface. There is a nice treatment in Igor Dolgachevs "Topics in classical algebraic geometry I" section 8.5 (including explicit rationalization, which is what you need).

Answer 5

You want positive integer solutions. Necessary conditions are that you can solve the first equation for c and the second for d. The discriminants must be squares, so

1+36*a^2-12*a+36*b^2-12*b = square1
1+36*a^2-12*a-36*b^2+12*b = square2

whereas one expects this to give an elliptic curve, it also allows for a quick and dirty computer search. Searching, I found quite a few integer solutions, but in the range 0 < a < 4001 and 0 < b < 4001 there was only one positive integer solution:

(a,b,c,d) = (2167,1020,2395,1912)