Problems & Puzzles:
Puzzle 79.- The
Chebrakov's - by two months - challenge
exists a 3x3 magic square composed only by squared numbers?"
According to the "Unsolved Problems of
Number Theory", R. K. Guy's book, p. 171, Martin LaBar
was who first proposed this question in 1984.
No one example of this kind of magic
squares is known up today.
Thanks to a kind email forwarded to me by Mutsumi
Suzuki (2/2/2000) it's supposed that at least one man in this planet already
knows exactly why no one example has been found. His name is Yuriy Chebrakov, a Russian Professor
who is prone to publish a paper (or a book) with a proof about the non
existence of the mentioned magic squares. But let's read to Chebrakov
directly through his email to Mr. Suzuki:
"I have very simple proofs that Magic Squares 3x3 from Fibonacci
and square numbers do not exist. The method how to prove it is adduced
in my book on Magic Squares... According to my plans I should finish
paper with title "Non-existing Magic Squares" to March 2000.
In particular, this paper will contain decision of the problem on Magic
Squares 3x3 from square numbers. Thus, readers of your web-site have
3 months to try themselves to solve this problem".(Underlining
is mine, CR)
With much less challenge than that I
have posted before some puzzles in my pages. But if the Chebrakov's
very nice playful-intriguing & puzzling attitude is not enough basis for this occasion, maybe
is interesting to know that:
a) Exist 4x4 magic squares of squares
b) The Randall's work on 3x3 semi-magical square of
squares has produced exactly the opposite possibility - according to
Randall - If exist infinite semi-magical squares of squares, maybe
one of them is absolutely magical!... Please click here and
to see the Randall's work and results.
Then, at least during two months or so,
we have the possibility of:
1) find a 3x3 magical
square of squares following the Randall's approach.
2) anticipate the Chebrakov's
(we strongly recommend to choose only one of them...)
Now, let's assume that the Chebrakov's
proof is valid, and that actually do not exist 3x3 magic squares of
squares at all. This means that we can not construct any 3x3 magic square
composed by nine (9) squared numbers. But what about 3x3 magic squares
with less than nine squared numbers?
I have produced a 3x3 magic square composed
by 7 squared numbers:
(primes - as rhapsodies - are in blue)
3) Exist 3x3 magic
squares composed by 8 squared numbers?
Finally, what about 3x3 magic squares
composed by squared primes? In this case, I guess I have an
extremely easy proof about the
impossibility of this kind of magic squares, that is to say, is impossible
to construct a 3x3 magic
squares consisting of 9 squared primes.
But I have produced the following - maybe
interesting - 3x3 magic squares using less than 9 squared primes:
A) This has 5 squared primes
(incidentally the four non-squared numbers - now in fucsia
- are also primes!)
B) The next one has 6 squared primes.
4) Can you prove that
does not exists a 3x3 magic square composed by squared primes?
5) Can you produce one 3x3 magic square with 7 or 8 squared primes?
6) Can you produce other magic squared type A)
7) Can you produce one 3x3 magic square with 6 squared primes and 3
Hints: Maybe can be useful (maybe not :-) you to see the
following Chebrakov's Web references:
Aale de Winkel is working in
this problem and has produced some preliminary results that you can see
Four years later (better late than
never!) J.C. Rosa produces some interesting result to this puzzle. He
has not gotten the supposedly existent Chebrakov's full proof but at
least a negative result for the 3x3 magic square of prime-squares.
About the question 4 of this puzzle
I have a proof that a 3x3 magic square composed by squared primes can
not exist. Here is my proof ( I hope that it is an easy proof ;- ) )
P1^2 P2^2 P3^2
P4^2 P5^2 P6^2
P7^2 P8^2 P9^2
If the square above is a magic
square we have , by example, the four pairs of equalities:
P5^2 - K2=P4^2 (2'
with K3=K1+K2 (5) and
If we have two equalities of
the following form :
P5^2 - K=Y^2 , we
can always write :
with X=p+q and Y=p-q , that
is to say : p=(X+Y)/2 and q=(X-Y)/2.
The sum of the equalities
(7) and (7 ' ) gives :
P5^2=p^2+q^2 (8 )
Therefore P5 is the
hypotenuse of a Pythagorean triangle.
We can write : P5=m^2+n^2
(9) ; p=2xmxn ; q=m^2-n^2 and K=2xpxq=4xmxmx(m^2-n^2).
If P5 is a prime number we
know the unicity of the equality (9) and we have seen above
that a magic square requires 4 different decompositions as the
equality (9). Thus a 3x3 magic square composed by 9 squared
primes can not exist.
Moreover , if P5 is prime
and if the square is magic then only the central row or the
central column or one of the two diagonals is composed by 3
squared primes . Thus if P5 is prime , the maximal number of
squared primes is equal to 6 .We have 2 possible kinds of
(o = squared prime ;
x=squared of a not prime )
Christian Boyer added this
one remark to this Jean-Claude Rosa’s proof:
sentence “If P5 is a prime number we know the unicity of the equality
(9)” is not completely true.
to change it by:
If P5 is a (4k+3) prime number, we know that the equality
(9) is impossible.
If P5 is a (4k+1) prime number, we know that the equality
(9) has only one solution.
this remark does not change the conclusion of the proof, a 3x3 magic
square composed by squares primes cannot exist.
If P5 is
a (4k+3) prime number, the maximum number of squared primes in the 3x3
magic square is 5.
If P5 is
a (4k+1) prime number, the maximum number of squared primes in the 3x3
magic square is 6.
below, you will find a slightly different proof:
going to through the centre cell c², and having two square integers
around the centre cell, is an integer solution of the equation:
x² + y² = 2c²
If c is a
4k+3 prime integer, then c cannot be a sum of two square integers. The
above equation is impossible.
If c is a
4k+1 prime integer, then c has only one way to be a sum of two square
integers c = a² + b².
implies that c² has only two different ways to be a sum of two square
c² = (2ab)² + (a² – b²)²
c² = (a² + b²)² + 0²
implies that 2c² has also only two solutions to be a sum of two square
2c² = (2ab + a² – b²)² + (2ab – a² + b²)²
2c² = (a² + b²)² + (a² + b²)²
second solution is impossible, all the cells of the magic square having
to be distinct.
- if the
centre cell of a 3x3 magic square is a squared prime c²=(4k+3)², then it
is impossible to have a magic line going through the centre and using
only squared integers.
- if the
centre cell of a 3x3 magic square is a squared prime c²=(4k+1)², then
there can be only one magic line going through the centre and using only
distinct squared integers:
c² = (a² + b²)²,
x² = (2ab + a² – b²)²,
y² = (2ab – a² + b²)².
cases, a 3x3 magic square of squared primes is of course impossible.
illustration is an analysis of the square type A).
centre cell is 197². The integer 197 being a (4k+1) prime number, there
is a unique way to write it as a sum of two square integers:
197 = 14² + 1²
two other cells of a line going through this centre are forced to be:
(2*14*1 + 14² – 1)² = 223²
(2*14*1 – 14² + 1)² = 167²
check that these numbers are used in the square A).
Landon W. Rabern has gotten and
published (Vol 4, (1),
2003) the following results about the magic square of squares. All these
results suppose that the magic square of squares is:
such that all the entries are coprime.
Theorem 1.1 All entries of the magic
square must be odd.
Theorem 1.2 The only prime divisors of
e are of the form p = 1(mod 4).
Theorem 1.3 If a prime p = 3; 5(mod 8)
divides a non-center entry then p
also divides the center and the other entry in that line.
Corollary 1.1 No prime p = 3(mod 8)
divides any entry.
Theorem 1.4 No prime p = 5(mod 8)
divides a middle-side entry.
Theorem 1.5 If a prime p = 3(mod 4)
divides a corner entry then it divides
the two middle-side entries that are not adjacent to the corner.
From these solid results Rabern
establish the following conjecture & challenge:
All of these properties taken together
severely restrict the possible placement
of primes that are not of the form p = 1(mod 8). Given these
might conjecture that if there is a solution, then all prime divisors of
are of the form p = 1(mod 8). This would greatly reduce the number of
It would also be interesting to
disprove this conjecture by proving the
opposite; namely, that any solution must have at least one entry with
divisor p = 5; 7(mod 8).
at Puzzle 288 with solutions of 5x5, 6x6 and 7x7 magic squares of squares"