Newsgroups: sci.math From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: Integers probem... Date: Sat, 22 Apr 1995 03:17:14 GMT In article <3n91n6$170@watson.math.niu.edu> rusin@washington.math.niu.edu (Dave Rusin) writes: >In article <3n8hlb$cmp@apopi.u-strasbg.fr>, >Christophe Lanuit wrote: >> Hi. I have some troubles with this little problem about integers : >> are they 3 integers (x>0,y>0,z>0) such as x^2+y^3=z^6 ??? >None. Dividing through by z^6 gives a rational point on the curve >Y^2=X^3+1 (Y=x/z^3, X=-y/z^2). You can, for example, use Cremona's >elliptic-curve program to find that this curve has rank 0 and torsion >Z/2Z, which is a long way of saying that the only rational point >on this curve is X=-1, Y=0. > The torsion group has order 6. Other rational points are (X, Y) = (0, +-1), (2, +-3), and the point at infinity. -- Peter L. Montgomery pmontgom@cwi.nl San Rafael, California Mathematically gifted, unemployed, U.S. citizen. Interested in computer architecture, program optimization, computer arithmetic, cryptography, compilers, computational mathematics. 17 years industrial experience. ============================================================================== From: kevin2003@delphi.com (Kevin Brown) Newsgroups: sci.math Subject: Re: Integers probem... Date: 23 Apr 1995 00:02:10 GMT CL = Christophe Lanuit CL> Hi. I have some troubles with this little problem about integers: CL> Are there 3 integers (x>0,y>0,z>0) such that x^2+y^3=z^6 ??? The equation x^2 + y^3 = z^6 has no solution in non-zero integers. To prove this, suppose there was such a solution. Clearly if any two of x,y,z have a common factor then they also share a common factor with the third, so we can divide until the three numbers x,y,z are coprime. Of all such solutions, let's consider the one with the smallest absolute value of z. The equation implies y^3 = (z^3)^2 - x^2, so we have y^3 = (z^3 + x)(z^3 - x) The two factors on the right side are mutually coprime, so they must each be cubes, meaning there are coprime integers a,b such that z^3 + x = a^3 and z^3 - x = b^3 Adding these two gives a^3 + b^3 = 2z^3. To prove that this equation has no non-zero integer solutions, assume a,b,z are the solution with the smallest value of z. Since a,b are coprime whereas the sum of their cubes is even, they must both be odd. This implies the existence of coprime integers u,v such that a=u+v and b=u-v. Substituting into a^3 + b^3 = 2z^3 and simplifying gives u (u^2 + 3v^2) = z^3 which proves that u and u^2 + 3v^2 must both be cubes. Thus, we have integers r,s such that u = r^3 and u^2 + 3v^2 = s^3 The right hand equation can be factored as (u + v sqr[-3]) (u - v sqr[-3]) = s^3 Following Euler's somewhat reckless path, we can take advantage of the fact that the ring of numbers of the form a + b sqr(-3) possesses unique factorization (where a,b are actually half-integers), and assert that each of the (coprime) factors on the left is a cube. Thus, we have coprime (half)integers g,h such that u + v sqr[-3] = (g + h sqr[-3])^3 = g (g^2 - 9h^2) + h (3g^2 - 3h^2) sqr[-3] from which it follows that u = g (g^2 - 9h^2) v = 3h (g^2 - h^2) Recalling that u = r^3, the left hand equation factors as r^3 = g (g+3h) (g-3h) (Notice that if g and h are really half-integers we can clear all fractions by multiplying through by 2^3, which still gives a cube on the left side.) Since g,h are coprime, each of the above three factors are coprime and therefore each must be cube. Thus, we have integers m,n,q such that g = m^3 g+3h = n^3 g-3h = q^3 which imply that n^3 + q^3 = 2m^3. But clearly m < g < r < z, so m < z, which contradicts our assumption that z is the smallest integer such that the double of its cube equals the sum of two cubes. ============================================================================== From: cet1@cus.cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: Integers probem... Date: 23 Apr 1995 15:20:40 GMT In article <9504221939591.kevin2003.DLITE@delphi.com>, kevin2003@delphi.com (Kevin Brown) writes: [snip] |> The equation x^2 + y^3 = z^6 has no solution in non-zero integers. Well, you aren't the first person in this thread to have missed the obvious x = +-3, y = -2, z = +-1. Only Peter Montgomery seems to have noticed this! [Reduction to x,y,z, pairwise coprime omitted] |> |> The equation implies y^3 = (z^3)^2 - x^2, so we have |> |> y^3 = (z^3 + x)(z^3 - x) |> |> The two factors on the right side are mutually coprime, so they must |> each be cubes, meaning there are coprime integers a,b such that |> |> z^3 + x = a^3 and z^3 - x = b^3 |> |> Adding these two gives a^3 + b^3 = 2z^3. Ah, but (z^3 + x) and (z^3 - x) need not be coprime: they could have a common factor of 2. In this case one of the factors is 4a^3 and one is 2b^3, so that we have 4a^3 + 2b^3 = 2z^3, or z^3 + (-b)^3 = 2a^3, which is the same equation as before with different variables... |> To prove that this equation |> has no non-zero integer solutions, assume a,b,z are the solution with |> the smallest value of z. But it has the trivial solution (1,1,1) [a torsion point on the elliptic curve]. For the first case, this gives rise only to a solution of the original with x=0, but in the second case we recover the solution with x=3. [snip] |> But clearly m < g < r < z, so |> m < z, which contradicts our assumption that z is the smallest integer |> such that the double of its cube equals the sum of two cubes. Not if m = g = r = z = 1, which is where you miss the torsion point. It's the usual problem: with induction you have to make sure you can start it off correctly, and for infinite descent you have to make sure that it really is turtles all the way down, and one of them isn't standing on itself! Chris Thompson Internet: cet1@cus.cam.ac.uk JANET: cet1@uk.ac.cam.cus ============================================================================== From: kevin2003@delphi.com (Kevin Brown) Newsgroups: sci.math Subject: integers problem Date: 26 Apr 1995 02:38:45 GMT KB> The equation x^2 + y^3 = z^6 has no solution in non-zero integers. CT = Chris Thompson CT> Well, you aren't the first person in this thread to have missed CT> the obvious x = +-3, y = -2, z = +-1. Only Peter Montgomery seems CT> to have noticed this! Yes, in fact (+-3k^3)^2 + (-2k^2)^3 = (+-k)^6 for all k. To be fair to the original poster, he specified positive integers, so this counter- example applies only to my thoughtless extension to NON-ZERO integers. Let me see if I can apply your corrections to tidy things up... KB> The equation implies...coprime integers a,b such that KB> a^3 + b^3 = 2z^3. CT> Ah, but (z^3 + x) and (z^3 - x) need not be coprime: they could CT> have a common factor of 2...so that we have...z^3 + (-b)^3 = 2a^3 CT> ...the first case...gives rise only to a solution of the original CT> with x=0, but in the second case we recover the solution with x=3. Yes, so in either case we have an equation of the form A^3 + B^3 = 2C^3, and if we take the trivial solutions {A=B=C} and {A=-B,C=0} we get one of the "obvious" solutions of the original equation {x=0, y=k^2, z=+-k} {x=+-k^3, y=-k^2, z=0} {x=+-3k^3, y=-2k^2, z=+-k} It only remains to show that an equation of the form a^3 + b^3 = 2z^3 has no solutions other than those with a=+-b. KB> Since a,b are coprime whereas the sum of their cubes is even, they KB> must both be odd. This implies the existence of coprime integers KB> u,v such that a=u+v and b=u-v. Here I should have specified {a not equal to +-b}, rather than just stating that a,b were coprime. Given this proviso, my previous proof holds. KB> ... clearly m < g < r < z, so m < z, which contradicts our KB> assumption that z is the smallest integer such that the double KB> of its cube equals the sum of two cubes. CT> Not if m = g = r = z = 1, which is where you miss the torsion point. If a,b are coprime then a=u+v and b=u-v where u and v are non-zero. Then r = z/((u^2 + 3v^2)^(1/3)) so clearly r < z, and it follows that m < z. I "missed the torsion point" when I neglected the trivial solutions a=+-b, for which either u or v equals zero, in which case u and (u^2 + 3v^2) are not coprime, so they need not both be cubes. So, with your corrections, we can say that the only solutions in non-zero integers of the equation x^2 + y^3 = z^6 are given by {x=+-3k^3, y=-2k^2, z=+-k} where k is any non-zero integer. Thus, there are no solutions in positive integers.