Two methods for finding 3x3 semi-magic squares of cubes
by Lee Morgenstern, May 2010.
(Second method corrected by Lee Morgenstern in April 2015, after remarks received from Tim S. Roberts)

Here is a way of searching for a 3x3 semi-magic square of cubes that extends the entries to twice the number of digits.

We previously searched for a solution using a list containing values which are the difference of two cubes in three different ways. This was the formulation.

A^3 B^3 C^3
D^3 E^3 F^3
G^3 H^3 I^3

C^3 - D^3  =  E^3 - I^3  =  G^3 - B^3
C^3 - H^3  =  E^3 - A^3  =  G^3 - F^3

If you just tried finding two triples where the C,E,G values already matched, then you didn't test all the possibilities.

Suppose you had two triples which didn't match

a^3 - b^3  =  c^3 - d^3  =  e^3 - f^3
g^3 - h^3  =  i^3 - j^3  =  k^3 - l^3

where a /= g, c /= i, e /= k

It could still be a solution if a/g = c/i = e/k. This is because each row can be scaled by a different cube which then forces them to match the pattern.

Suppose gcd(m,n) = 1 and n/m = a/g = c/i = e/k, then the smallest scaling would be

(am)^3 - (bm)^3  =  (cm)^3 - (dm)^3  =  (em)^3 - (fm)^3
(gn)^3 - (hn)^3  =  (in)^3 - (jn)^3  =  (kn)^3 - (ln)^3

Since am = gn, cm = in, em = kn, we have a solution.

A = jn
B = fm
C = am = gn
D = bm
E = cm = in
F = ln
G = em = kn
H = hn
I = dm

If the list of triples contained cubes of 6-digit numbers, then using the scaling method above could potentially find a solution using the cubes of 12-digit numbers.

(I think that the previous search only looked at primitive triples. All triples need to be checked to have a reasonable chance of finding something).

Here is yet another way of searching for a 3x3 semi-magic square of cubes that might be better than my scaling-triple method. It involves more scalings so this can produce solutions with even larger entries than ever before.

In the following, assume all variables are cubes.

Formulation

A B C
D E F
G H I

A + B = I + F
A + D = I + H

A + G = E + F
A + C = E + H

Find two taxicabs

a + b = c + d
e + f = g + h

such that ag = ce.

Cross multiply so that A and I match.

ae + be = ce + de  -->  A + B = I + F
ae + af = ag + ah  -->  A + D = I + H

Find another two taxicabs

i + j = k + l
m + n = p + q

such that ip = km.

Cross multiply so that A and E match.

im + jm = km + lm  -->  A + G = E + F
im + in = ip + iq  -->  A + C = E + H

Cross multiply so that A, F and H match

aeim + beim = ceim + deim  -->  A + B = I + F
aeim + afim = agim + ahim  -->  A + D = I + H
aeim + aejm = aekm + aelm  -->  A + G = E + F
aeim + aein = aeip + aeiq  -->  A + C = E + H

This requires

al = di  and
eq = hm.

We then have the semi-magic square

A = aeim
B = beim
C = aein
D = afim
E = aeip = aekm --> ip = km
F = aelm = deim --> al = di
G = aejm
H = aeiq = ahim --> eq = hm
I = agim = ceim --> ag = ce

aeim    beim       aein
afim  aeip=aekm  aelm=deim
aejm  aeiq=ahim  agim=ceim