Recently I posted here that if a+b and a^2-a*b+b^2 are
both 3 times squares, then a parameterized solution for
the Diophantine equation
(1) 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)
This gives infinitely many solutions, but not all solutions,
as Hans Havermann has pointed out to me in private email.
Since several sequences in OEIS are concerned with the
Diophantine equation (1), let me say a little more about it.
First, if (a,b,c) is a solution of (1) then so is
(k^2*a,k^2*b,k^3*c), for any k. So let us say that two solutions
(a,b,c) and (a',b',c') are equivalent if there is a non-zero
rational number k such that (a',b',c')=(k^2*a,k^2*b,k^3*c).
This is an equivalence relation and each equivalence class has a
reduced (or primitive) representative (a,b,c) such that
a,b,c are integers and the largest integer n such that
n^2|a and n^2|b and n^3|c is n=1. All this works just as
well if a,b,c are allowed to be rational numbers and we can
pass from a rational solution of (1) to the primitive integer
solution in the same equivalence class by using the proper
value for k. Thus we may as well be looking for all rational
points on the curve x^3+y^3=c^2. This is actually an elliptic
curve, and as we vary c we get a family of ellipic curves.
So, we are really looking for all rational points on the elliptic
surface x^3+y^3=z^2. Is is not reasonable to expect a complete
parametric solution, but in the case where a+b and a^2-a*b+b^2
are both three times a square such a solution is given below.
Theorem: (a,b,c) is a rational solution to (1) a^3+b^3=c^2 in
which a+b and a^2-a*b+b^2 are both 3 times a rational square,
if and only if there exist rational numbers u and v such that
a = u*(3*u^2*v^4-1+6*u*v^2),
(2) b = -u*(3*u^2*v^4-1-6*u*v^2),
c = 6*v*u^2*(3*u^2*v^4+1).
Some examples:
solution (a,b,c) to (1) values for (u,v) in (2)
(11, 37, 228) (16, 1/8)
(37, 11, 228) (3, 2/3)
(1, 2, 3) (3/4, 2/3)
(100, 200, 3000) (75, 1/15)
(71, -23, 588) (1, 2)
(23/2, 1/2, -39) (1/2, -2)
I won't give a proof of the theorem here, but substituting (2) into
(1) gives an identity so that direction is just a computation. Also,
notice that replacing u by k^2*u and v by v/k changes (a,b,c) given
by (2) to (k^2*a,k^2*b,k^3*c) so we can easily move within any
equivalence
class of solutions. Hence it suffices to show that there exist
rationals
u,v such that the right sides of (2) give some (a',b',c') equivalent to
the given (a,b,c). I'll leave it at that for now.
Some sequences related to the Diophantine equation (1) are:
A099806, A099807, A099808, A099809, A098970, A099426.
Jim Buddenhagen
------------------------------------------------------------------------
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
------------------------------------------------------------------------
Ed Pegg Jr wrote:
> Kirk Bresniker notes that for a^3 + b^3 = c^2, if a and b
> are prime, then for a** Seems like it should be extendable and provable.
(I'm changing a and b to p and q, since that's what I have in my notes,
and if I try to change them all I'll probably miss some.)
Here's a proof. First, if p=2, then q^3 + 8 = c^2. According to
http://mathworld.wolfram.com/MordellCurve.html
this equation is known to have only 4 solutions in integers, namely
(q,c) = (-2,0), (1,3), (2,4), (46,312). None of these have q prime and
p**= 3. (5) implies r = p = q, contradicting
p