From: ejones@hooked.net (Earle D. Jones) Newsgroups: sci.math Subject: FNTLT (Fermat's Next-to-Last Theorem) Date: Sun, 24 Mar 1996 11:47:06 -0800 1. How many solutions to: a^2 + b^2 = c^4 (a, b, c integers) Example: 7^2 + 24^2 = 5^4 2. How many solutions to: a^2 + b^2 = c^4 (a, b, c integers, a = b+1) Example: 120^2 + 119^2 = 13^4 Have fun! earle ===== -- ..no sig is good sig.. ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: FNTLT (Fermat's Next-to-Last Theorem) Date: 26 Mar 1996 22:37:31 GMT In article , Earle D. Jones wrote: >1. How many solutions to: > > a^2 + b^2 = c^4 (a, b, c integers) > > Example: 7^2 + 24^2 = 5^4 This equation has infinitely many solutions, e.g. a= 7 d^2 b=24 d^2 c=5 d But there are more solutions than this. The equation may be written r*rbar = c^4 where r=a+bi and rbar=a-bi are elements of Z[i]. If a and b are relatively prime, then any common divisor of r and rbar would also divide r+rbar = 2a and -i(r-rbar) = 2b, and hence would divide 2. Using the unique factorization in Z[i], we see r must be of the form r = u * s^4 for some unit u = 1, -1, i, -i and some s = x + iy in Z[i]. (I should consider the possibility that 1+i divides r, but that would make c even, whence a^2+b^2 = 0 mod 4, which can only happen if a and b are also even, hence not coprime.) Of course any such r does indeed lead to a solution: r*rbar would be u*ubar * (s*sbar)^4, which is the fourth power of a rational integer. Thus we parameterize the solutions with relatively prime a and b as a = x^4 - 6x^2y^2 + y^4 b = 4x^3y - 4xy^3 c = x^2+y^2 (as well as the permutations of a and b, and sign changes of any of them.) The poster's solution comes from x=1, y=2. >2. How many solutions to: > > a^2 + b^2 = c^4 (a, b, c integers, a = b+1) > > Example: 120^2 + 119^2 = 13^4 I suspect the poster has already found what is essentially the unique nontrivial solution to this problem. If there are integer solutions (a,b,c), then x=c and y=a+b are integers with y^2=2x^4-1, which is an elliptic curve and thus as Siegel has shown has only finitely many integral points. Siegel's proof is not effective, although this can be reduced to a Thue equation, which is effectively solvable although I don't have a program to do it (I think one is included in the latest version of KASH but I don't have that version.) Here's what to do. First note that if x is even, we'd have y^2=-1 mod 4, a contradiction. Next, write the equation as y^2+1=2x^4 and factor over Z[i] as (y+i)(y-i)=2x^4. As in the solution to part 1, any common divisor of y+i and y-i divides their difference 2i, from which it's easy to check (since 2 does not divide x) that y+i = (1+i) u r^4 y-i = (1-i) u^(-1) rbar^4 for some unit u (=1, -1, i, or -i) and some element r=a+bi of Z[i]. Subtracting and considering the various cases for u leads in all cases to an equation of the form a^4-4a^3b-6a^2b^2+4ab^3+b^4 = +- 1 The left side is an irreducible homogeneous form of degree greater than 2, so that this is a Thue equation, and hence (once again) has only finitely many solutions. Maple has a Thue solver but it doesn't know when it has found all solutions; still it covers its search space pretty quickly (I had it search for solutions with |b|<10^30 in a few seconds) and so I have reasonable confidence that the small solutions it does find (|a|. |b| <=3 ) are the only ones. I also thought to solve the equation rationally -- if the curve had rank zero, this would make it easy to find the integral solutions. Unfortunately, it has rank 1, so that approach doesn't work. Here are the details. We seek rational points on the elliptic curve y0^2 = 2 x0^4 - 1. Following Cassel's book we reduce this to canonical form using the rational point (x0=1, y0=1). This leads ultimately to the substitutions 2 - 94 x - 8 x + 13 x y + 184 y - 832 x0 := ---------------------------------------------- 2 64 - 520 x - 44 x + 280 y + x y and 4 3 2 2 2 y0 := - (5736 x - 87952 x + 321216 x - 239 x y 2 2 2 + 48528 y x - 85248 x y - 24976 x y + 978944 + 49216 y 3 + 3456 y - 433152 y) / 2 2 / (64 - 520 x - 44 x + 280 y + x y) / Here's the result when I asked Maple to > factor(simplify(y0^2-2*x0^4+1)); 2 3 2 32 (1728 y + 832 - 3376 x - 239 x ) (- x - 8 x + y ) 3 / 2 4 (- 25 x + 56 + 6 y) / (64 - 520 x - 44 x + 280 y + x y) / Setting the first factor to zero simplifies (x0, y0) to the constant point (13, -239). Setting the third factor to zero likewise yields only one point (1, -1). Other than at those two points, we have y0^2-2*x0^4+1=0 only when y^2 = x^3 + 8 x. (Conversely, whenever y^2 = x^3 + 8 x, we get points with y0^2=2*x0^4-1 except with (x=8, y=24), which makes the above denominator vanish. There are two other problem points over suitable extensions of Q as well.) So apart from a few points, we see that the original quartic curve is rationally isomorphic to the curve y^2 = x^3+8x. This is a curve with one torsion point of order 2 at (0,0), AND an infinite cyclic component generated by (1, 3) or (1, -3). Substituting the "smallest" points (excluding (8,24)) on this curve into our formulas for (x0, y0) give the solutions (1,1) (-1,1) (-1,-1) (-13,-239) (so far so good) and then (1525/1343, -2750257/1803649) (-2165017/2372159, -3503833734241/5627138321281) "and so on". I'm guessing the original poster didn't expect solutions like a= 1061652293520/5627138321281, b=-4565486027761/5627138321281, c= 2165017/2372159 existed, although he asked explicitly for integer solutions. dave ============================================================================== From: fc3a501@rzaixsrv1.uni-hamburg.de (Hauke Reddmann) Newsgroups: sci.math Subject: Re: FNTLT (Fermat's Next-to-Last Theorem) Date: 27 Mar 1996 13:01:54 GMT Earle D. Jones (ejones@hooked.net) wrote: : 1. How many solutions to: : a^2 + b^2 = c^4 (a, b, c integers) : Example: 7^2 + 24^2 = 5^4 Infinite: a pythagorean triple is a=2kl,b=k**2-l**2,c=k**2+l**2. If k and l are the first of a p.t. themselves, k**2+l**2 is a square, so c a biquadrate. : 2. How many solutions to: : a^2 + b^2 = c^4 (a, b, c integers, a = b+1) : Example: 120^2 + 119^2 = 13^4 The latter is the only one, AFAIK. CF "Game,Set,Math" by Ian Stewart with an xref to Ljunggren. -- Hauke Reddmann <:-EX8 fc3a501@math.uni-hamburg.de UP AGAIN! HOORAY! fc3a501@rzaixsrv1.rrz.uni-hamburg.de IF NOT reddmann@chemie.uni-hamburg.de SCIENCE ONLY ============================================================================== From: cet1@cus.cam.ac.uk (Chris Thompson) Newsgroups: sci.math Subject: Re: FNTLT (Fermat's Next-to-Last Theorem) Date: 30 Mar 1996 23:46:31 GMT In article <4j9rjb$3md@muir.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) writes: |> |> >2. How many solutions to: |> > |> > a^2 + b^2 = c^4 (a, b, c integers, a = b+1) |> > |> > Example: 120^2 + 119^2 = 13^4 |> |> I suspect the poster has already found what is essentially the unique |> nontrivial solution to this problem. If there are integer solutions |> (a,b,c), then x=c and y=a+b are integers with y^2=2x^4-1, which is |> an elliptic curve and thus as Siegel has shown has |> only finitely many integral points. Siegel's proof is not effective, |> although this can be reduced to a Thue equation, which is effectively |> solvable although I don't have a program to do it (I think one is included |> in the latest version of KASH but I don't have that version.) This particular equation has a (short) section to itself in Richard Guy's UPINT (section D6). To quote Ljunggren has shown that the only solutions of y^2 = 2x^4 - 1 in positive integers are (1,1) and (239,13) but his proof is difdicult. Mordell asks if it is possible to find a simple or elementary proof. Whether Steiner & Tzankis have simplified the solution may be a matter of taste: they use the theory of linear forms in the logarithms of algebraic numbers. and goes on to consider y^2 = Dx^4 + 1 for other values of D. Incidentally, this 1-parameter class of Thue equations is one of those which Voutier & Chen[? - I am having difficulty reading my notes here :-(] have a uniform result for using hypergeometric methods: for !D| > 2 there is at most one solution in positive integers. Chris Thompson Email: cet1@cam.ac.uk ============================================================================== Date: Thu, 28 Mar 1996 13:33:09 +1100 From: ftww@cs.su.oz.au (Fred the Wonder Worm) To: rusin@vesuvius.math.niu.edu, rusin@vesuvius.math.niu.edu Subject: Re: FNTLT (Fermat's Next-to-Last Theorem) In article <4j9rjb$3md@muir.math.niu.edu> you write: > [ ... ] Siegel's proof is not effective, >although this can be reduced to a Thue equation, which is effectively >solvable although I don't have a program to do it (I think one is included >in the latest version of KASH but I don't have that version.) Yes, the latest version of KASH can solve Thue equations. (Although there appears to be a bug with solving for negative numbers.) I used Magma to solve this instead (it incorporates most of the KANT code). >Subtracting and considering the various cases for u leads in all cases >to an equation of the form > a^4-4a^3b-6a^2b^2+4ab^3+b^4 = +- 1 >The left side is an irreducible homogeneous form of degree greater than 2, >so that this is a Thue equation, and hence (once again) has only finitely >many solutions. The solutions found were: [ -3, 2 ], [ -2, -3 ], [ -1, 0 ], [ 0, -1 ], [ 0, 1 ], [ 1, 0 ], [ 2, 3 ], [ 3, -2 ] which would confirm your statement. Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. ------------------------------------------------------------------------------- ============================================================================== Subject: Re: FNTLT (Fermat's Next-to-Last Theorem) To: rusin@math.niu.edu (Dave Rusin) Date: Thu, 28 Mar 1996 16:40:08 +1000 (EST) From: "Fred the Wonder Worm" > Thanks for the tip. I have to learn to think of using Magma for other than > group theory problems. The appropriate functions involved are ThueObject, ThueEval, ThueSolve and ThueSolveInexact. Because of the bug I mentioned before, the ThueSolve function only works for non-negative values. Also, ThueObject was called just Thue at some stage - it will depend on which version of Magma you have. > t:=ThueObject(x^4-4*x^3-6*x^2+4*x+1); > print ThueSolveInexact(t,1); generated the solutions I mentioned previously. Cheers, Geoff. ------------------------------------------------------------------------------- Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@cs.su.oz.au | Gameplayer by vocation. -------------------------------------------------------------------------------