Problems & Puzzles:
Puzzles
Puzzle 287.
Multimagic prime squares
For sure you know already the interesting and nice site of
Christian Boyer about the multimagic squares. If not, maybe is a good idea
you review it before going on with this puzzle.
In any case the very basic concept is this one:
A magic square is bimagic (or 2-multimagic) if it
remains magic after each of its numbers have been squared. By extension, a
square is P-multimagic if it remains magic after each of its numbers have
been replaced by their k-th power (for k=1, 2, ..., to P).
Some basic facts about P-multimagic squares are:
- The first bimagic historically produced is an 8x8 square
constructed in 1890 by
G. Pfefferman.
- No bimagic is possible (using consecutive numbers) for a nxn square for n = 3, 4, 5 &
6. For n=7 the impossibility hasn't been completely established, but is
believed to be impossible.
- The smallest orders n currently known of P-multimagic
squares are:
| P |
n |
| 2 |
8, 9, 10, 11,16 |
| 3 |
12, 32, 64, 128 |
|
4 |
256,
512 |
|
5 |
729,
1024 |
| 6 |
4096 |
Well, this week Christian and me crossed emails about Mr. Chebrakov (See our
Puzzle 79). In some moment during this interchange, I
asked him (13/10/04):
One quick question: Is possible to get multimagic
prime-squares?
He responded (13/10/04):
Hmmmm, I do not know. A very VERY difficult question.
Imagine that, today, nobody has succeeded to construct a bimagic square of
distinct numbers which is smaller than the well-known 8x8 bimagic squares
of consecutive numbers. And "distinct" is far less limited than "prime"!
A 3x3 bimagic square using distinct integers (or of course prime integers)
is proved to be impossible. My 4x4 best result is this semi-bimagic
square:
| 9 |
55 |
105 |
36 |
| 69 |
100 |
21 |
15 |
| 28 |
49 |
19 |
109 |
| 99 |
1 |
60 |
45 |
4 bimagic rows, 4 bimagic columns, but only one magic diagonal S1=205,
S2=15427.
You can find in my site a 6x6 square constructed by Pfeffermann in 1894:
6 bimagic rows, 6 bimagic columns, two magic diagonals (but not bimagic).
I replied (13/10/04):
What I wanted to know is if there is some theoretical
impossibility in order to get such kind of objects (prime-bimagic
squares). May I suppose from your response that you are not aware of such
impossibility?
In his turn Christian responded (14/10/04):
Yes, I can't see any theoretical impossibility of a nxn
bimagic square of prime numbers, with n>=4. But impossible for n<4.
Thinking about this problem, I have created this morning this nice (and
probably first) semi-bimagic square of prime numbers:
| 29 |
293 |
641 |
227 |
| 277 |
659 |
73 |
181 |
| 643 |
101 |
337 |
109 |
| 241 |
137 |
139 |
673 |
You can check that:
1) The 16 used numbers are prime integers, and are
distinct.
2) The 4 rows and 4 columns have the same sum S1=1190.
3) And when these 16 prime numbers are squared, the 4 rows and 4 columns
have the same sum S2=549100.
4) But the 2 diagonals are not magic or bimagic.
Who will be the first to create a bimagic square of
prime numbers, including two bimagic diagonals?
It is a VERY difficult problem. Remember that nobody has
succeeded to solve the "easier" problem to create a bimagic square of
distinct integers, the square being smaller than 8x8.
Enough!...
Questions:
1. Improve the Boyer 4x4 prime
semi-bimagic square (at least one bimagic
diagonal).
2. Get a (full) prime-bimagic square
(what is the minimal nxn order for this target?)
Preliminary results related to question 2 came from Luke Pebody
and J. C. Rosa.
Both proved that is impossible to construct 4x4 bimagic squares of
distinct numbers.
Here is the J. C. Rosa's demonstration:
Square I
a b c d
e f g h
i j k l
m n o p
Square II
a^2 b^2 c^2 d^2
e^2 f^2 g^2 h^2
i^2 j^2 k^2 l^2
m^2 n^2 o^2 p^2
If the square I is magic we have (*), by
example, the equality: a+m=h+l (1)
If the square II is magic we have the same
thing :
a^2+m^2=h^2+l^2 (2)
The equalities (1) and (2) give : a*m=h*l
(3)
(remark: if a,m,h,l are prime numbers the
equality (3) is impossible )
The equalities (1) and (3) give the
following equation :
m^2-(h+l)*m+h*l=0
This equation has two solutions: m=h or m=l. Therefore a 4x4 bimagic square with 16
distinct numbers is impossible.
_________
(*) This supposes - for example - the validity of the Bergholt's formula
for 4x4 magic squares; otherwise (1) needs a demonstration.
The Luke's proof is this one:
There are no bimagic squares of distinct integers for 4x4.
I do not use the lines through E,H,I or L in the diagram below, but I
do use both bimagic diagonals.
Proof:
Let
ABCD
EFGH
IJKL
MNOP
be a magic square.
Then:
A+B+C+D=S1
M+N+O+P=S1
A+F+K+P=S1
D+G+J+M=S1
B+F+J+N=S1
C+G+K+O=S1
Adding the top 3, and subtracting the bottom 3:
2(A+P)=2(G+J).
Therefore A+P=G+J.
Similarly, by the bimagic property, A^2+P^2=G^2+J^2. Therefore {A,P}={G,J}.
QED
***
Christian Boyer sends (July 2005) the following line: "Look
at Puzzle 288 where the two above proofs are mentioned in a mathematical
article published in 2005"
...
Later, on Nov. 2006, he added:
I am
happy to announce the first known BIMAGIC square of primes. And its
order is also prime: 11. See this
page
by Boyer.
Reminder. A bimagic square of order n is a nxn magic square
remaining magic after each of its n² numbers have been squared.
|
137 |
131 |
317 |
47 |
5 |
457 |
541 |
359 |
467 |
353 |
683 |
|
401 |
277 |
239 |
647 |
23 |
421 |
229 |
181 |
7 |
419 |
653 |
|
463 |
269 |
701 |
59 |
157 |
257 |
563 |
557 |
179 |
191 |
101 |
|
593 |
311 |
379 |
503 |
197 |
83 |
53 |
521 |
149 |
619 |
89 |
|
307 |
617 |
397 |
241 |
571 |
661 |
109 |
107 |
79 |
127 |
281 |
|
373 |
443 |
29 |
587 |
383 |
61 |
19 |
409 |
631 |
389 |
173 |
|
73 |
11 |
607 |
433 |
613 |
577 |
263 |
97 |
227 |
313 |
283 |
|
43 |
599 |
151 |
199 |
509 |
487 |
223 |
163 |
293 |
691 |
139 |
|
673 |
37 |
113 |
271 |
193 |
31 |
601 |
431 |
331 |
337 |
479 |
|
67 |
233 |
103 |
439 |
499 |
251 |
547 |
659 |
491 |
41 |
167 |
|
367 |
569 |
461 |
71 |
347 |
211 |
349 |
13 |
643 |
17 |
449 |
Characteristics:
-121
distinct prime integers, and more precisely
121 consecutive prime integers <= 701, only excluding 2, 3, 523,
641, 677
-bimagic square (11 rows, 11 columns, 2 diagonals) with magic sums
S1=3497, S2=1578251
It is the first solved problem on the 10 open problems published in
my Math Intelligencer article, Spring 2005.
But
it means also that 9 problems are still unsolved. A lot of work
remains to be done, including "small" unsolved problems:
- 3x3
magic square of squares (only semi-magic are known)
- 4x4 magic square of cubes (only semi-magic are known)
- 5x5 bimagic square (only semi-bimagic are known)
***
Update (April 2007)
C. Boyer wrote:
...questions 1 and 2 of puzzle 287 are still unsolved:
-nobody knows a better 4x4 example
-nobody knows a 4x4, 5x5,… 10x10 bimagic square of primes.
However, you will see
here
a 10x10 nearly bimagic square of primes.
***
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