Problems & Puzzles: Puzzles

Puzzle 79.- The Chebrakov's - by two months - challenge

"Does 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

 13^2 42^2 1^2 36^2 44^2 9^2 22^2 27^2 6^2 19^2 48^2 23^2 33^2 32^2 21^2 26^2

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 here 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 proof.

(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:

 565^2 23^2 222121 289^2 425^2 527^2 373^2 360721 205^2

(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!)

 19^2 7^2 313 193 241 17^2 13^2 433 11^2

B) The next one has 6 squared primes.

 65737 31^2 223^2 151^2 197^2 54817 167^2 76657 109^2

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 non-squared primes

***
Hints: Maybe can be useful (maybe not :-) you to see the following Chebrakov's Web references:
1) http://forum.swarthmore.edu/epigone/historia_matematica/blexkhehsner
2) http://at.yorku.ca/cgi-bin/amca/cacf-11

Solution

Aale de Winkel is working in this problem and has produced some preliminary results that you can see at:www.adworks.myweb.nl/Gen_Magic/MagicSequences.html

***

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+K1=P3^2      (1)

P5^2 - K1=P7^2      (1' )

P5^2+K2=P6^2      (2)

P5^2 - K2=P4^2      (2' )

P5^2+K3=P1^2      (3)

P5^2 - K3=P9^2      (3' )

P5^2+K4=P8^2      (4)

P5^2 - K4=P2^2      (4' )

with K3=K1+K2   (5)  and  K4=K1+K3=2xK1+K2   (6)

If we have two equalities of the following form :

P5^2+K=X^2

P5^2 - K=Y^2  , we can always write :

P5^2+K=(p+q)^2  (7)

P5^2 - K=(p-q)^2  (7' )

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 squares:

o    x    o                   o    o    o

o    o    o       or         o    o    x

x    o    x                   x    x    o

(o = squared prime ; x=squared of a not prime )

***

Christian Boyer added this (30/10/04)

I have one remark to this Jean-Claude Rosa’s proof:

The sentence “If P5 is a prime number we know the unicity of the equality (9)” is not completely true.

We have 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.

Of course 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.

Just below, you will find a slightly different proof:

----------------------------

A line 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 with x≠y≠c

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.

It implies that c² has only two different ways to be a sum of two square numbers:

-        c² = (2ab)² + (a² – b²)²

-        c² = (a² + b²)² + 0²

It implies that 2c² has also only two solutions to be a sum of two square numbers:

-        2c² = (2ab + a² – b²)² + (2ab – a² + b²)²

-        2c² = (a² + b²)² + (a² + b²)²

The second solution is impossible, all the cells of the magic square having to be distinct.

The conclusion:

- 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.

In both cases, a 3x3 magic square of squared primes is of course impossible.

An illustration is an analysis of the square type A).

Its 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

Then the 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

You can 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:

a2 b2 c2
d2 e2 f2
g2 h2 s2

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 restrictions, one
might conjecture that if there is a solution, then all prime divisors of all entries
are of the form p = 1(mod 8)
. This would greatly reduce the number of possibilities.

It would also be interesting to disprove this conjecture by proving the
opposite; namely, that any solution must have at least one entry with prime
divisor p = 5; 7(mod 8).

***

"Look at Puzzle 288 with solutions of 5x5, 6x6 and 7x7 magic squares of squares" (C. Boyer)

***

Arkadiusz Wesolowski wrote on Dec 29, 2018:

Q6:

Here are another 3x3 magic squares (type A).

 11^2 23^2 433 673 19^2 7^2 17^2 193 601

 23^2 1801 1777 2617 37^2 11^2 31^2 937 47^2

 29^2 47^2 1993 2833 41^2 23^2 37^2 1153 2521

Q5:

As a small digression I remark that there exists a simple way for generating an infinite number of 3x3 magic squares with 5 perfect squares.

Let x = -1 + ((sqrt(2)/2)*((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + (3 + 2*sqrt(2))^n + (3 - 2*sqrt(2))^n)/2 for some integer n >= 1, and let:

a = 17*x^4 + 44*x^3 + 34*x^2 + 10*x + 1,

b = 9*x^2 + 6*x + 1,

c = 10*x^4 + 28*x^3 + 23*x^2 + 8*x + 1,

d = 2*x^4 + 8*x^3 + 11*x^2 + 6*x + 1,

e = 9*x^4 + 24*x^3 + 22*x^2 + 8*x + 1,

f = 16*x^4 + 40*x^3 + 33*x^2 + 10*x + 1,

g = 8*x^4 + 20*x^3 + 21*x^2 + 8*x + 1,

h = 18*x^4 + 48*x^3 + 35*x^2 + 10*x + 1,

i = x^4 + 4*x^3 + 10*x^2 + 6*x + 1.

Then the magic square is

 a b c d e f g h i

Examples:

1)      n = 1

 7753 13^2 4753 35^2 65^2 85^2 3697 91^2 697

2)      n = 2

 11441977 85^2 6779473 1189^2 2465^2 3277^2 5372977 3485^2 710473

The value of x can be found by using the recurrence formula: a(n) = 6*a(n-1) - a(n-2) + 4 with a(0) = 0 and a(1) = 4. If we let a(n) = 6*a(n-1) - a(n-2) - 4 with a(0) = 2 and a(1) = 6, then we can also produce an infinite number of 3x3 magic squares with 5 perfect squares using the following equalities.

Let x = 1 + ((sqrt(2)/2)*((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + (3 + 2*sqrt(2))^n + (3 - 2*sqrt(2))^n)/2 for some integer n >= 1, and let:

a = 17*x^4 - 44*x^3 + 34*x^2 - 10*x + 1,

b = 9*x^2 - 6*x + 1,

c = 10*x^4 - 28*x^3 + 23*x^2 - 8*x + 1,

d = 2*x^4 - 8*x^3 + 11*x^2 - 6*x + 1,

e = 9*x^4 - 24*x^3 + 22*x^2 - 8*x + 1,

f = 16*x^4 - 40*x^3 + 33*x^2 - 10*x + 1,

g = 8*x^4 - 20*x^3 + 21*x^2 - 8*x + 1,

h = 18*x^4 - 48*x^3 + 35*x^2 - 10*x + 1,

i = x^4 - 4*x^3 + 10*x^2 - 6*x + 1.

Then the magic square is

 a b c d e f g h i

Examples:

1)      n = 1

 13693 17^2 7693 35^2 85^2 115^2 6757 119^2 757

2)      n = 2

 12612301 89^2 7364461 1189^2 2581^2 3451^2 5958661 3649^2 710821

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