Analysis of a quadratic diophantine equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:41:11Z http://mathoverflow.net/feeds/question/43489 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation Analysis of a quadratic diophantine equation apples 2010-10-25T06:48:47Z 2010-10-31T22:03:42Z <p>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.</p> <p>So far, I only really have managed to deduce the following things:</p> <p>$2b(3b-1) + d(3d-1) = a(3a-1) +b(3b-1)$</p> <p>$2b(3b-1) = c(3c-1) - d(3d-1)$</p> <p>$2b(3b-1) = (c-d)(3(c+d)-1)$</p> <p>Any ideas on where to go from here would be greatly appreciated. Thanks!</p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/43490#43490 Answer by Robin Chapman for Analysis of a quadratic diophantine equation Robin Chapman 2010-10-25T06:59:51Z 2010-10-25T15:17:32Z <p>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 $6$.</p> <p>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.</p> <p><strong>Added</strong> 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 <strong>rational</strong> solutions to the original pair of equations.</p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/43575#43575 Answer by rita the dog for Analysis of a quadratic diophantine equation rita the dog 2010-10-25T20:43:43Z 2010-10-25T20:43:43Z <p>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</p> <p>1+36*a^2-12*a+36*b^2-12*b = square1<br> 1+36*a^2-12*a-36*b^2+12*b = square2</p> <p>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 &lt; a &lt; 4001 and 0 &lt; b &lt; 4001 there was only one positive integer solution:</p> <p>(a,b,c,d) = (2167,1020,2395,1912)</p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/43577#43577 Answer by David Lehavi for Analysis of a quadratic diophantine equation David Lehavi 2010-10-25T20:59:51Z 2010-10-25T21:05:49Z <p>As Robin and Fedor observed the variety in question is a quartic Del Pezzo surface. There is a nice treatment in <a href="http://www.math.lsa.umich.edu/~idolga/topics.pdf" rel="nofollow">Igor Dolgachevs "Topics in classical algebraic geometry I"</a> section 8.5 (including explicit rationalization, which is what you need).</p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/44255#44255 Answer by Franz Lemmermeyer for Analysis of a quadratic diophantine equation Franz Lemmermeyer 2010-10-30T16:48:49Z 2010-10-30T16:48:49Z <p>Completing the squares and setting $X = 6a-1$, $Y = 6b-1$, $Z = 6c-1$ and $W = 6d-1$ we find $$X^2 + Y^2 = Z^2 + 1, \quad X^2 - Y^2 = W^2 - 1.$$</p> <p>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.</p> <p>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, $$(1) \qquad \qquad Y^2 = 4tu(t-u)(t-2u) + 1.$$</p> <p><b> 1. Brute Force </b> A direct search for points on this surface yields several solutions; the solutions with $t &lt; (2-\sqrt{2}\,)u$ 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<br> $(t,u,a,b,c,d) = (51, 2506, 2051177, 415877, 2092912, 2008575)$.</p> <p>This suggests that the diophantine problem has infinitely many solutions in positive integers. Proving this conjecture seems to be difficult, however.</p> <p><b> 2. Elliptic Surfaces</b></p> <p>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 $(t,y) = (0,0), (0,-1), (u,0), (u,-1), (2u,0), (2u,-1)$.</p> <p>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 $2$.</p> <p>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.</p> <p><b>3. The Fibonacci connection. </b> 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. </p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/44272#44272 Answer by John R Ramsden for Analysis of a quadratic diophantine equation John R Ramsden 2010-10-30T19:57:10Z 2010-10-31T22:03:42Z <p>For this system one can find a general rational parametrization and then N&amp;S conditions for integer solutions.</p> <p>Adding the pair:</p> <p>$x^2 + y^2 = z^2 + 1$</p> <p>$x^2 - y^2 = t^2 - 1$</p> <p>gives:</p> <p>$2 x^2 = z^2 + t^2$</p> <p>which has a general rational parametrization (GPR):</p> <p>$(z + t)/2 = (v^2 - 1) x / (v^2 + 1)$</p> <p>$(z - t)/2 = 2 v x / (v^2 + 1)$</p> <p>Adding these gives an expression for z and plugging this back in the first of the original pair then gives:</p> <p>$(y/x)^2 - (1/x)^2 = 4 v (v^2 - 1)/(v^2 + 1)^2$ ( $= 4 R$ say)</p> <p>which has GPR:</p> <p>$y/x, 1/x = (L^2 + R)/L, (L^2 - R)/L$</p> <p>and replacing $u := L (v^2 + 1)$ (to give an obvious simplification) yields a general rational solution of the original pair as:</p> <p>$D = u^2 - v (v^2 - 1)$</p> <p>$D x = u (v^2 + 1)$</p> <p>$D y = u^2 + v (v^2 - 1)$</p> <p>$D z = u (v^2 + 2 v - 1)$</p> <p>$D t = u (v^2 - 2 v - 1)$</p> <p>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.</p> <p>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:</p> <p>$b L (b^2 - c^2) = (L - 2) a^2 c$</p> <p>Then two cases arise, depending on the parity of L.</p> <h2>Case 1 L odd</h2> <p>We show that this is impossible (given the other constraints of the problem).</p> <p>If $L$ is odd then $(L, L - 2) = 1$ and thus for some integer $n$ we have:</p> <p>$a^2 c, b(b^2 - c^2) = L n, (L - 2) n$</p> <p>Multiplying the equations for $z$ and $t$ by $2 a b$, and plugging the above pair into the result gives:</p> <p>$2 a b z = a^2 (L - 2) + 2 b^2 L$</p> <p>$2 a b t = a^2 (L - 2) - 2 b^2 L$</p> <p>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:</p> <p>$2 z = A (L - 2) / B + B (2 L / A)$</p> <p>$2 t = A (L - 2) / B - B (2 L / A)$</p> <p>For z, t to be integers we require $A (L - 2) / B$ and $B (2 L / A)$ to both odd or both even.</p> <p>If they are both odd then A and $2 L / A$ must be both odd, which is impossible.</p> <p>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.</p> <p>So that leaves us with ..</p> <h2>Case 2 L even</h2> <p>Denoting $m := L / 2$ for convenience, we must how have for some integer $n$ :</p> <p>$a^2 c, b(b^2 - c^2) = m n, (m - 1) n$ [*]</p> <p>which, as in Case 1, implies:</p> <p>$2 z = A (m - 1) / B + B (2 m / A)$</p> <p>$2 t = A (m - 1) / B - B (2 m / A)$</p> <p>Again $A (m - 1) / B$ and $B (2 m / A)$ must be either both odd or both even..</p> <p>Both odd leads to the same contradiction as Case 1 as it requires $A$ and $2 m / A$ both odd.</p> <p>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 $m - 1$).</p> <p>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.</p> <p>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!</p>