Fw: Divisible by 12: a^3 + b^3 = c^2
James Buddenhagen
jbuddenh at earthlink.net
Mon Oct 25 00:34:18 CEST 2004
My apologies if you see this twice, it never showed up here in
San Antonio, Texas
JB
----- Original Message -----
From: "James Buddenhagen" <jbuddenh at earthlink.net>
To: "Pfoertner, Hugo" <Hugo.Pfoertner at muc.mtu.de>; "'Ed Pegg Jr'" <edp at wolfram.com>
Cc: "seqfan" <seqfan at ext.jussieu.fr>; "James Buddenhagen" <jbuddenh at earthlink.net>
Sent: Saturday, October 23, 2004 5:56 PM
Subject: Re: Divisible by 12: a^3 + b^3 = c^2
> Ed Pegg, based on numerical information from Kirk Bresniker,
> asks whether a,b prime and a^3 + b^3 = c^2 implies that
> c is divisible by 12. Hugo Pfoertner, Ralf Stephan, an
> myself provided additional numerical evidence and comments.
> Finally, Paul C. Leopardi proved a lemma that is almost the
> desired result, but unless I misunderstood, needed an
> additional hypotheses.
>
> Here is additional information which amounts to a proof
> and some parameterized solutions, but some details omitted.
>
> Lemma 1: if gcd(a,b)=1 then gcd(a+b,a^2-a*b+b^2)=1 or 3.
>
> proof: Since (a+b)^2-(a^2-a*b+b^2) = 3*a*b, any prime p<>3
> which divides both a+b and a^2-a*b+b^2 will divide a or b and
> a+b hence it will divide both a and b contrary to gcd(a,b)=1.
> Similarly 3^k for k>1 cannt be a common divisor.
>
> Lemma 2: If gcd(a,b)=1 and a^3 + b^3 = c^2 then either
> (1) a+b and a^2-a*b+b^2 are both squares or
> (2) a+b and a^2-a*b+b^2 are both 3 times squares.
>
> proof: a^3+b^3 = (a+b)*(a^2-a*b+b^2) = c^2 and lemma 1.
>
> Now complete parametric solutions can be given for both
> (1) and (2). The technique to find these is the standard
> method of finding all rational points on a quadratic curve
> when you know one point, by intersecting with lines of
> rational slope through the point. To get integers we need
> to write the slope as a quotient of integers and multiply by
> appropriate squares to eliminate denominators. I omit details.
>
> But the results are:
>
> (1) if a+b and a^2-a*b+b^2 are both squares then a paramterized:
> solution of a^3+b^3=c^2 is:
>
> a = -4*n*(-2*n+m)*(n^2-m*n+m^2)
> b = (m-n)*(m+n)*(7*n^2-4*m*n+m^2)
> c = (n^2-4*m*n+m^2)*(13*n^4-14*m*n^3+6*m^2*n^2-2*m^3*n+m^4)
>
> Note: a+b = (n^2-4*m*n+m^2)^2 and
> a^2-a*b+b^2 = (13*n^4-14*m*n^3+6*m^2*n^2-2*m^3*n+m^4)^2
>
> Note 2: m and n are integers so certainly a and b cannot be
> prime.
>
> Note 3: it is possible in this case for a and b to be relatively
> prime and c not divisible by 12.
>
> (2) if a+b and a^2-a*b+b^2 are both 3 times squares then a
> parameterized solution for a^3+b^3=c^2 is:
>
> a = 3*m^4-n^4+6*m^2*n^2
> b = -3*m^4+n^4+6*m^2*n^2
> c = 6*m*n*(3*m^4+n^4)
>
> Note: a+b = 12*m^2*n^2
> a^2-a*b+b^2 = 3*(3*m^4+n^4)^2,
> and each is 3 times a square.
>
> Note 2: if m and n have the same parity then a and b are
> not relatively prime, so for relatively prime a,b
> exactly one of m,n is even. In this case
> a+b is 48 times a square and c is divisible by 12.
>
> Theorem: if a,b are prime and a^3 + b^3 = c^2, then
> 48 divides a+b with the quotient a square,
> and 12 divides c.
>
> proof: since a and b are both prime we are not in case (1) above,
> see note 2 of that case. So we are in case (2) above and
> the result follows by Note 2 of that case.
>
>
> The parameterization in (2) allows us to very quickly find
> prime numbers a and b such that a^3 + b^3 = c^2. Here are
> the first 30, sorted by c, extending Hugo Pfoertner's data.
> Following Paul Leopardi's notation let d=sqrt((a+b)/48).
>
> m n a b c c/12 a+b d
>
> 1 2 11 37 228 19 48 1
> 6 5 8663 2137 812340 67695 10800 15
> 8 7 28703 8929 4935504 411292 37632 28
> 10 7 56999 1801 13608420 1134035 58800 35
> 8 11 44111 48817 14218512 1184876 92928 44
> 11 8 86291 6637 25354032 2112836 92928 44
> 10 11 87959 57241 29463060 2455255 145200 55
> 7 16 16931 133597 48880608 4073384 150528 56
> 13 14 246011 151477 135516108 11293009 397488 91
> 9 20 54083 334717 194057640 16171470 388800 90
> 16 11 367823 3889 223078944 18589912 371712 88
> 17 14 552011 127717 412662012 34388501 679728 119
> 14 23 457511 786697 763311948 63609329 1244208 161
> 15 22 571019 735781 764539380 63711615 1306800 165
> 16 25 765983 1154017 1409359200 117446600 1920000 200
> 23 16 1586531 38557 1998370272 166530856 1625088 184
> 21 26 1915163 1662229 3408412644 284034387 3577392 273
> 22 29 2437751 2446777 5397667572 449805631 4884528 319
> 15 38 16139 3882661 7650577620 637548135 3898800 285
> 21 34 2305883 3811669 8224333236 685361103 6117552 357
> 26 29 4074743 2747449 9401817516 783484793 6822192 377
> 25 32 3963299 3716701 10658164800 888180400 7680000 400
> 19 40 1296563 5634637 13456391280 1121365940 6931200 380
> 28 31 5440991 3600097 14413082712 1201090226 9041088 434
> 25 44 4683779 9836221 32471808600 2705984050 14520000 550
> 22 47 2238023 10591849 34633513596 2886126133 12829872 517
> 32 37 9682703 7139569 35661291456 2971774288 16822272 592
> 30 41 8681639 9473161 38787516180 3232293015 18154800 615
> 29 44 8142803 11395309 44940252984 3745021082 19538112 638
> 29 46 8321723 13032949 52820789196 4401732433 21354672 667
>
> We can also find silly prime curiosities such as:
>
> let a,b be the 99 digit primes
> 516311827796740838771674147960359381895640671834628612875993558021288236937751709944319900868251071
> and
> 164347159141325505114613844644012165100482223256876273832159859778089839875893509442524862953510657
> respectively. Then a^3 + b^3 is the square of an integer and
> a/b = 3.14159265358979323846259..
>
> Jim Buddenhagen
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