I appreciate raw fodder for thought. You have provided me with some critical tidbits in the past, so my thanks are overdue.

I am simply agape that solutions of

a^3 + b^3 = c^3 + d^3

a^3 + b^3 = c^3 + d^3 = e^3 + f^3 with a,b,c,d,e, and f being distinct positive integers?

70^3 + 560^3 = 98^3 + 552^3 = 315^3 + 525^3

This one is fun to figure out:

are so commonplace.>

315^3 + 525^3 = 198^3 + 552^3 = 70^3 + 560^3 = 175959000,

whereas

255^3 + 414^3 = 228^3 + 423^3 = 167^3 + 436^3 = 87539319.

The final expression you wrote can't be correct. Look at the rightmost digits:

...9 + ...0 = ...5 + ...5 = ...2 + ...8 = ...5 + ...5.

It's still not right. I checked the *three* least significant digits in each expression. All are of the form ...000 except the second member:

...275^3 + ...775^3 = ...250.

912655621660656000= 545275^3 + 908775^3

545275 = 545,000 + 275 and 908775 = 908,000 + 775.

275^3 = 20796875 and 775^3 = 465484375.

Hence

545275^3 = 1000s + 275^3 = 1000s + 20796875 = 1000s' + 875,

where s is an undetermined whole number and s' = s + 20796;

and

908775^3 = 1000t + 775^3 = 1000t + 465484375 = 1000t' + 375,

where t is some undetermined whole number and t' = t + 465484.

So 545275^3 + 908775^3 = 1000(s' + t') + 875 + 375 = 1000u + 250,

where u = s' + t' + 1.

I was thinking that maybe the numbers are too big (too many digits) for your spreadsheet, just as they are too big for my hand calculator to calculate exactly. If your spreadsheet works with 64-bit arithmetic, then the biggest number it can handle correctly (assuming one bit for a ± sign) is 2^63 ≈ 9.2 x 10^18. But 908775^3 is smaller than that and three of the other numbers that it apparently calculated correctly are greater than 908775, so that explanation can't be right.

I worked around this problem in my Fortran code by expanding the cubes using (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.

Here's my listing (CG's software screws up the formatting, so you'll have to fix that yourself and redo the prettyprinting or imagine it):

program Taxicab4
! ---Declare
parameter (n = 10000)
integer*4 i, j, k, m, n, kount, least, most
integer*8 term1(n), term2(n), term3(n), term4(n) parameter (LDO = 16)
! ---Initialize
OPEN (UNIT=LDO, FILE='OUTPUT', STATUS='UNKNOWN') do m = 1, n
least = mod(m, 1000)
most = (m - least)/1000
term1(m) = most**3
term2(m) = 3*most**2*least
term3(m) = 3*most*least**2
term4(m) = least**3
enddo
kount = 0
! ---Main loop
do k = 1, n
do j = 1, k-1
do i = 1, j-1
if (term1(i) + term1(k) .eq. 2*term1(j)) then if (term2(i) + term2(k) .eq. 2*term2(j)) then
if (term3(i) + term3(k) .eq. 2*term3(j)) then if (term4(i) + term4(k) .eq. 2*term4(j)) then write (LDO,1) i, k, j
kount = kount + 1
endif
endif
endif
endif
enddo
enddo
enddo
write (LDO,2) kount
1 format(I5, '^3 +', I5, '^3 = 2 ×', I5, '^3') 2 format(I6, ' examples found.')
end

Suppose that a, b, and c are positive integers satisfying

(1) a^3 + b^3 = 2c^3,

and that there is no smaller set of such numbers. If any two of them had a common factor d > 1, they all would. Then a/d, b/d, and c/d would be smaller numbers and would satisfy the same Eq. (1), so no pair of a, b, and c can have a common factor.

By Fermat's Little Theorem, a^3 ≡ a (mod 3); b^3 ≡ b (mod 3); c^3 ≡ c (mod 3), so Eq. (1) implies that a + b ≡ 2c (mod 3), or

(2) a + b + c ≡ 0 (mod 3).

Hence the residues of a, b, and c modulo 3 are either all equal or all different.

Since the RHS of (1) is even, a and b must both be odd. (If both were even they would have the common factor 2.) An odd integer has residue 1, 3, or 5 modulo 6, so a and b must take one of the following five forms:

(i) a = 6j + 1, b = 6k+ 1;

(ii) a = 6j + 3, b = 6k + 1;

(iii) a = 6j + 5, b = 6k + 1;

(iv) a = 6j + 5, b = 6k + 3;

(v) a = 6j+ 5, b = 6k + 5,

where j and k are integers, or the corresponding forms with the residues interchanged, which are equivalent. (The case a = 6j + 3, b = 6k + 3 can be eliminated because then a and b would have a common factor 3.)

Expanding the LHS of Eq. (1) in these five cases we find that it can be written respectively as

(i) a^3 + b^3 = 6m + 2, so c^3 = 3l + 1;

(ii) a^3 + b^3 = 6m + 4, so c^3 = 3l + 2;

(iii) a^3 + b^3 = 6m, so c^3 = 3l;

(iv) a^3 + b^3 = 6m + 2, so c^3 = 3l + 1;

(v) a^3 + b^3 = 6m + 4, so c^3 = 3l + 2,

where l is an integer that depends on the values of j and k. So by Fermat's theorem in these five cases c takes the form

(i) c = 3l + 1;

(ii) c = 3l + 2;

(iii) c = 3l ;

(iv) c = 3l + 1;

(v) c = 3l + 2,

respectively, where l depends on j and k.

If c is even the RHS of Eq. (1) contains a factor 16. Since a and b are odd, it must be possible to write them as a = 16s + r and b = 16t − r, where the residue r is chosen from 1, 3, 5, 7, 9, 11, 13, 15, or the corresponding forms with the sign of r reversed.

Fermat's Last Theorem (his *big* theorem) says that for n greater than 2 there are no positive integers a, b, c satisfying a^n + b^n = c^n. For n = 3 this equation is the same as Eq. (1), except that the factor 2 on the RHS is missing. There are elementary proofs of the theorem for that special case, including one due to Euler (http://2000clicks.com/mathhelp/Numb...), so I looked at it to see if the technique he used works here.

Since a and b are odd, if we set

(3) 2u = a + b; 2v = a − b,

u and v are whole numbers satisfying a = u + v and b = u − v. They have opposite parity and no common factor except 1. Then Eq. (1) becomes

(4) u(u^2 + 3v^2) = c^3.

The greatest common divisor (GCD) of u and u^2 + 3v^2 must be either 1 or 3. (If it were any other number, it would divide both u and v.) Suppose it is 1. Since c is a cube, both u and u^2 + 3v^2 must be cubes (otherwise they would have to have a common factor):

(5) u = s^3; u^2 + 3v^2 = t^3.

If the cube of t equals the sum of a square and three times a second square, then t must also have that form:

(6) t = p^2 +3q^2 .

This is not easy to prove. Using the “complex product identity,” it is straightforward to show that if t = p^2 + 3q^2 and t' = p'^2 + 3q'^2, then t × t' = p"^2 + 3q"^2, where

(7) p" = pp' – 3qq' and q" = pq' + p'q.

It follows that any power of t has the same form; in particular,

(8) t^3 = p"'^2 + 3q"'^2

where p"' = p^3 – 9pq^2 and q"' = 3qp^2 – 3q^3.

It's easy to construct numerical examples illustrating this. Thus,

13^3 = 13^2 + 3·26^2, and 13 = 1^2 + 3·2^2;
16^3 = 32^2 + 3·32^2, and 16 =2^2 + 3·2^2;
21^3 = 63^2 + 3·42^2, and 21 = 3^2 + 3·4^2;
49^3 = 48^2 + 3·192^2, and 49 = 1^2 + 3·4^2;
52^3 = 260^2 + 3·156^2 and 52 = 5^2 + 3·3^2, etc.

But this doesn't prove that numbers with this property must exist. What is needed is a proof, starting from Eq. (8), with p"', q"', and t known, that p and q satisfying Eq. (6) can always be found. Euler apparently didn't prove it, but later mathematicians have, so I will apply the result to p"' = u and q"' = v.

Factor the expression for p"':

(9) p^3 – 9pq^2 = p(p + 3q)(p – 3q) = u = s^3.

If p, p + 3q, and p – 3q had a factor in common, it would divide u and v, so each is separately a cube:

(10) p + 3q = a'^2; p – 3q = b'^2; p = c'^2.

Hence

(11) a'^3 + b'^3 = 2p = 2c'^3.

This has the same form as Eq. (1). But by construction a', b', and c' are smaller than a, b, and c, contradicting the initial assumption, so it's impossible.

If the GCD of u and u^2 + 3v^2 is 3, then 3 is a factor of u, so u = 3w for some integer w, but 3 does not divide v because they have no common factor. Then

(12) u(u^2 + 3v^2) = 3w[(3w)^2 + 3v^2] = 9w(v^2 + 3w^2) = c^3.

By the same reasoning as above, where now v takes the place of u and w takes that of v, there must be numbers x and y such that v = x^3 – 9xy^2 and w = 3yx^2 – 3y^3. Hence

(13) 9w = 27(yx^2 – y^3) = 27y(y + x)(y – x)

is a cube, from which it follows that y + x = a"^3, y – x = b"^3, and y = c"^3 are all cubes, and

(14) a"^3 + b"^3 = 2y = 2c"^3.

If a^3 + b^3 = 2*c^3

then 4*a^3 + 4*b^3 = 8*c^3 = (2c)^3

with

4*a^3 + 4*b^3 =

(a+b)*((a+b)^2 + 3*(a-b)^2)

and

Let f = a + b and g = a – b.

f must divide 2c: 2c = ef.

So f(f^2 + 3g^2) = e^3f^3.

Cancelling f, this becomes

3(e^3 - 1)f^2 = (3g)^2.

So f divides 3g: 3g = fh.

Since f > g, this is only possible if (i) h = 1; (ii) h = 2; or (iii) h = 3.

(i) 3a - 3b = a + b, or a = 2b. They have a common factor.

(ii) 2a - 2b = a + b, or a = 3b. They have a common factor.

(iii) b = 0. Trivial.

you mean this one:

(2*c)^3 = (2*c)*((2*c)^2 + 3*0)?>

f must divide 2c: 2c = ef. ....>