8x8 and 9x9 magic squares of cubes.
8x8 and 9x9 magic squares of 4th
powers.
8x8 and 9x9 magic squares of 5th powers.
See also the Magic
squares of cubes general page
In this 8x8 semi-magic square of cubes using distinct positive integers from 1 to 65, by Lee Morgenstern, only 57 is missing. It is not a magic square: the diagonals are not magic.
|
65 3 |
64 3 |
9 3 |
12 3 |
6 3 |
23 3 |
7 3 |
3 3 |
|
2 3 |
29 3 |
63 3 |
25 3 |
45 3 |
13 3 |
51 3 |
33 3 |
|
20 3 |
48 3 |
19 3 |
62 3 |
17 3 |
15 3 |
52 3 |
34 3 |
|
28 3 |
22 3 |
38 3 |
32 3 |
61 3 |
37 3 |
41 3 |
44 3 |
|
10 3 |
11 3 |
60 3 |
40 3 |
27 3 |
42 3 |
30 3 |
53 3 |
|
43 3 |
21 3 |
26 3 |
46 3 |
24 3 |
59 3 |
16 3 |
50 3 |
|
8 3 |
31 3 |
18 3 |
35 3 |
58 3 |
54 3 |
14 3 |
49 3 |
|
55 3 |
47 3 |
4 3 |
39 3 |
5 3 |
36 3 |
56 3 |
1 3 |
After a long and difficult search, the first known 8x8 magic squares of cubes were constructed by Walter Trump, Germany, in August and September 2008. Here are his two magic squares of cubes.
|
11 3 |
9 3 |
15 3 |
61 3 |
18 3 |
40 3 |
27 3 |
68 3 |
|
3 3 |
56 3 |
12 3 |
25 3 |
49 3 |
38 3 |
65 3 |
16 3 |
|
21 3 |
34 3 |
64 3 |
57 3 |
32 3 |
24 3 |
45 3 |
14 3 |
44 3 |
4 3 |
8 3 |
48 3 |
54 3 |
66 3 |
13 3 |
9 3 |
|
|
38 3 |
3 3 |
58 3 |
8 3 |
66 3 |
2 3 |
46 3 |
10 3 |
27 3 |
67 3 |
21 3 |
30 3 |
6 3 |
11 3 |
33 3 |
63 3 |
|
|
63 3 |
31 3 |
41 3 |
30 3 |
13 3 |
42 3 |
39 3 |
50 3 |
10 3 |
52 3 |
51 3 |
45 3 |
50 3 |
36 3 |
15 3 |
47 3 |
|
|
37 3 |
51 3 |
12 3 |
6 3 |
54 3 |
65 3 |
23 3 |
19 3 |
58 3 |
19 3 |
24 3 |
17 3 |
60 3 |
20 3 |
37 3 |
53 3 |
|
|
47 3 |
36 3 |
43 3 |
33 3 |
29 3 |
59 3 |
52 3 |
4 3 |
23 3 |
14 3 |
61 3 |
62 3 |
28 3 |
43 3 |
32 3 |
31 3 |
|
|
55 3 |
53 3 |
20 3 |
49 3 |
25 3 |
16 3 |
5 3 |
56 3 |
41 3 |
26 3 |
57 3 |
46 3 |
18 3 |
1 3 |
59 3 |
40 3 |
|
|
1 3 |
62 3 |
26 3 |
35 3 |
48 3 |
7 3 |
60 3 |
22 3 |
64 3 |
2 3 |
42 3 |
39 3 |
5 3 |
55 3 |
34 3 |
35 3 |
In order to reach the fourth power, with:
we can construct this semi-magic square, sum Sn = uvw:
|
(aei)n |
(afi)n |
(agj)n |
(ahj)n |
(bgk)n |
(bhk)n |
(bel)n |
(bfl)n |
|
(bei)n |
(bfi)n |
(bgj)n |
(bhj)n |
(agk)n |
(ahk)n |
(ael)n |
(afl)n |
|
(cej)n |
(cfj)n |
(cgi)n |
(chi)n |
(dgl)n |
(dhl)n |
(dek)n |
(dfk)n |
|
(dej)n |
(dfj)n |
(dgi)n |
(dhi)n |
(cgl)n |
(chl)n |
(cek)n |
(cfk)n |
|
(cfl)n |
(cel)n |
(chk)n |
(cgk)n |
(dhj)n |
(dgj)n |
(dfi)n |
(dei)n |
|
(dfl)n |
(del)n |
(dhk)n |
(dgk)n |
(chj)n |
(cgj)n |
(cfi)n |
(cei)n |
|
(afk)n |
(aek)n |
(ahl)n |
(agl)n |
(bhi)n |
(bgi)n |
(bfj)n |
(bej)n |
|
(bfk)n |
(bek)n |
(bhl)n |
(bgl)n |
(ahi)n |
(agi)n |
(afj)n |
(aej)n |
With:
we get the (big) smallest solution using this method:
|
797094 |
27214934 |
27047964 |
39107564 |
63503364 |
91816964 |
2842424 |
97048344 |
|
2134584 |
72880664 |
72433524 |
104728724 |
23713284 |
34286084 |
1061414 |
36239574 |
|
2718524 |
92818044 |
40300334 |
58268634 |
54067664 |
78174264 |
2401284 |
81986564 |
|
2738964 |
93515924 |
40603344 |
58706744 |
53664174 |
77590874 |
2383364 |
81374724 |
|
81692594 |
2392674 |
77288964 |
53455364 |
88820564 |
61430964 |
61810184 |
1810344 |
|
82306824 |
2410664 |
77870084 |
53857284 |
88157724 |
60972524 |
61348914 |
1796834 |
|
36098564 |
1057284 |
34420014 |
23805914 |
69221384 |
47875584 |
110265044 |
3229524 |
|
96670724 |
2831364 |
92175624 |
63751424 |
25848494 |
17877594 |
41174924 |
1205964 |
But, again, the diagonals can't be magic with this method.
Who will be the first to construct a 8x8 magic square of fourth powers? Or prove that it is impossible?
Who will be the first to construct a 8x8 magic square of fifth powers? Or prove that it is impossible?
9x9 magic
squares of cubes
9x9 magic squares of fourth powers
9x9 magic squares of fifth powers
Lee Morgenstern found a method to construct 9x9 semi-magic squares of cubes (or of any nth-powers). If the two equations (9.1) (9.2) are true:
then this square is a semi-magic square of nth-powers, with magic sum Sn = uv:
|
(aj)n |
(bj)n |
(cj)n |
(dk)n |
(ek)n |
(fk)n |
(gl)n |
(hl)n |
(il)n |
|
(ak)n |
(bk)n |
(ck)n |
(dl)n |
(el)n |
(fl)n |
(gj)n |
(hj)n |
(ij)n |
|
(al)n |
(bl)n |
(cl)n |
(dj)n |
(ej)n |
(fj)n |
(gk)n |
(hk)n |
(ik)n |
|
(bm)n |
(cm)n |
(am)n |
(ep)n |
(fp)n |
(dp)n |
(hq)n |
(iq)n |
(gq)n |
|
(bp)n |
(cp)n |
(ap)n |
(eq)n |
(fq)n |
(dq)n |
(hm)n |
(im)n |
(gm)n |
|
(bq)n |
(cq)n |
(aq)n |
(em)n |
(fm)n |
(dm)n |
(hp)n |
(ip)n |
(gp)n |
|
(cr)n |
(ar)n |
(br)n |
(fs)n |
(ds)n |
(es)n |
(it)n |
(gt)n |
(ht)n |
|
(cs)n |
(as)n |
(bs)n |
(ft)n |
(dt)n |
(et)n |
(ir)n |
(gr)n |
(hr)n |
|
(ct)n |
(at)n |
(bt)n |
(fr)n |
(dr)n |
(er)n |
(is)n |
(gs)n |
(hs)n |
Some months later, I constructed -not using the above method- the first known 9x9 magic square of cubes. It uses the 81 first cubes, from 13 to 813. An added property: the 9 rows (and 3 columns) are magic when the numbers are not cubed, S1=369.
It should be the smallest possible "normal" magic square of cubes, using the first consecutive cubes, 8x8 or smaller "normal" magic squares of cubes seeming impossible.
|
53 3 |
2 3 |
68 3 |
72 3 |
19 3 |
62 3 |
28 3 |
49 3 |
16 3 |
|
9 3 |
70 3 |
40 3 |
12 3 |
24 3 |
30 3 |
78 3 |
50 3 |
56 3 |
|
27 3 |
36 3 |
8 3 |
69 3 |
34 3 |
58 3 |
39 3 |
17 3 |
81 3 |
|
1 3 |
11 3 |
23 3 |
47 3 |
71 3 |
76 3 |
45 3 |
43 3 |
52 3 |
|
75 3 |
63 3 |
33 3 |
25 3 |
42 3 |
22 3 |
21 3 |
74 3 |
14 3 |
|
65 3 |
73 3 |
7 3 |
41 3 |
6 3 |
66 3 |
54 3 |
31 3 |
26 3 |
|
51 3 |
13 3 |
35 3 |
4 3 |
55 3 |
15 3 |
60 3 |
77 3 |
59 3 |
|
3 3 |
57 3 |
64 3 |
32 3 |
80 3 |
29 3 |
48 3 |
10 3 |
46 3 |
|
61 3 |
20 3 |
79 3 |
67 3 |
38 3 |
5 3 |
44 3 |
18 3 |
37 3 |
It is also possible to construct a 9x9 magic square of cubes using numbers from 03 to 803. As it is in the previous square, its 9 rows are again magic when the numbers are not cubed, S1=360.
|
0 3 |
11 3 |
14 3 |
57 3 |
76 3 |
60 3 |
49 3 |
50 3 |
43 3 |
|
69 3 |
10 3 |
45 3 |
40 3 |
70 3 |
38 3 |
2 3 |
21 3 |
65 3 |
|
9 3 |
73 3 |
13 3 |
36 3 |
41 3 |
62 3 |
34 3 |
72 3 |
20 3 |
|
54 3 |
28 3 |
55 3 |
35 3 |
5 3 |
66 3 |
17 3 |
22 3 |
78 3 |
|
75 3 |
58 3 |
26 3 |
64 3 |
59 3 |
3 3 |
31 3 |
32 3 |
12 3 |
|
24 3 |
8 3 |
80 3 |
33 3 |
47 3 |
61 3 |
52 3 |
4 3 |
51 3 |
|
39 3 |
15 3 |
63 3 |
23 3 |
7 3 |
19 3 |
71 3 |
67 3 |
56 3 |
|
42 3 |
53 3 |
30 3 |
79 3 |
1 3 |
44 3 |
27 3 |
68 3 |
16 3 |
|
48 3 |
74 3 |
46 3 |
29 3 |
18 3 |
37 3 |
77 3 |
6 3 |
25 3 |
Remarks on the Morgenstern's 9x9 method
Using his method above for cubes (n=3), we cannot directly use the smallest Taxicab(3, 3, 3) numbers:
because the 81 generated integers would not be distinct. If my research is correct, the smallest solution using his method and producing 81 distinct integers is:
generating the square:
|
4 3 |
12 3 |
104 3 |
161 3 |
460 3 |
483 3 |
561 3 |
594 3 |
627 3 |
|
23 3 |
69 3 |
598 3 |
231 3 |
660 3 |
693 3 |
68 3 |
72 3 |
76 3 |
|
33 3 |
99 3 |
858 3 |
28 3 |
80 3 |
84 3 |
391 3 |
414 3 |
437 3 |
|
24 3 |
208 3 |
8 3 |
200 3 |
210 3 |
70 3 |
648 3 |
684 3 |
612 3 |
|
30 3 |
260 3 |
10 3 |
720 3 |
756 3 |
252 3 |
144 3 |
152 3 |
136 3 |
|
108 3 |
936 3 |
36 3 |
160 3 |
168 3 |
56 3 |
180 3 |
190 3 |
170 3 |
|
416 3 |
16 3 |
48 3 |
567 3 |
189 3 |
540 3 |
551 3 |
493 3 |
522 3 |
|
702 3 |
27 3 |
81 3 |
609 3 |
203 3 |
580 3 |
304 3 |
272 3 |
288 3 |
|
754 3 |
29 3 |
87 3 |
336 3 |
112 3 |
320 3 |
513 3 |
459 3 |
486 3 |
It is the smallest possible square using the method. And it can't be magic: I have checked that no arrangement of this square allows to get at least one magic diagonal.
Using his method for fourth powers (n=4), we cannot directly use the two smallest Taxicab(4, 3, 3) numbers:
because the 81 generated integers would not be distinct, a problem on two couples of integers: 7*34 = 17*14 = 238, and 28*9 = 12*21 = 252. If I am right, the smallest solutions (=smallest possible S4) producing 81 distinct integers use the smallest Taxicab(4, 3, 3) number and the smallest Taxicab(4, 3, 4) number:
Combining these equations, we would be able to generate 4 squares, but the second equation cannot be used with (A) = (B) = (C), or (B) = (C) = (D), because three couples of integers would not be distinct: 4*45 = 12*15 = 180, 7*32 = 28*8 = 224, and 7*45 = 21*15 = 315.
Using (A) = (B) = (D), or (A) = (C) = (D), all the integers are distinct. Here is the square using (A) = (B) = (D).
|
124 |
694 |
814 |
2804 |
8404 |
11204 |
5164 |
7314 |
12474 |
|
1604 |
9204 |
10804 |
3014 |
9034 |
12044 |
364 |
514 |
874 |
|
1724 |
9894 |
11614 |
214 |
634 |
844 |
4804 |
6804 |
11604 |
|
1844 |
2164 |
324 |
7774 |
10364 |
2594 |
7654 |
13054 |
5404 |
|
8514 |
9994 |
1484 |
9454 |
12604 |
3154 |
1364 |
2324 |
964 |
|
10354 |
12154 |
1804 |
1684 |
2244 |
564 |
6294 |
10734 |
4444 |
|
6214 |
924 |
5294 |
7004 |
1754 |
5254 |
13924 |
5764 |
8164 |
|
6754 |
1004 |
5754 |
13444 |
3364 |
10084 |
6674 |
2764 |
3914 |
|
12964 |
1924 |
11044 |
6444 |
1614 |
4834 |
7254 |
3004 |
4254 |
The other square, using (A) = (C) = (D), has of course exactly the same S4.
Who will be the first to construct a 9x9 magic square of fourth powers? Or prove that it is impossible?
For the fifth power, nobody knows any Taxicab(5, 3, 3) number u:
After an exhaustive search done in March/April 2003, Duncan Moore did not find any solution u < 17716^5 = 1.745 * 10^21. His search on various Taxicab/Cabtaxi numbers can be found in his webpage at http://homepage.ntlworld.com/duncan-moore/taxicab. It means that, today, we can't construct a 9x9 semi-magic square of 5th powers using the Morgenstern's method.
Who will be the first to construct a 9x9 magic square of fifth powers? Or prove that it is impossible?
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