Every Puzzle find his appropriate puzzler... The Jaime Ayala original puzzle must have been published around January 5, 2001, according to WON plate 87. Then, this contribution by Arina Bator comes just 25 years after!!! (CR).

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Arina Bator wrote on May 12, 2026:

I am writing regarding Puzzle 124. Although I have not managed to find a complete solution, I do not know how far others have progressed, and I thought my partial results might be of interest.

To begin with, I would like to present my “best” result so far: a pair of magic squares. In the first square, exactly 6 entries satisfy all conditions (i.e. they are palindromic primes), and additionally 2 entries are prime but not palindromic. In the second square, obtained via the transformation m=2*n+1, there are again 6 entries satisfying all required conditions (palindromic primes).

Base magic square:

39180535053508193    39280524752508293    36380815351808363

35480905350908453    38280625052608283    41080344754308113

40180434753408203    37280725352708273     37380715051708373

Transformed square ( m=2*n+1):

78361070107016387   78561049505016587    72761630703616727

70961810701816907    76561250105216567   82160689508616227

80360869506816407   74561450705416547   74761430103416747

In my notation, bold numbers are neither palindromes nor primes, while underlined numbers are primes but not palindromes. (Qualification: 12/18, CR)

I started from the fact that constructing a 3×3 magic square can be reduced to the existence of three 3-term arithmetic progressions with common difference r, such that the first terms of these progressions themselves form a 3-term arithmetic progression with common difference d.

Based on this, I searched among pairs (n, 2*n+1) where both n and 2*n+1 are palindromic primes, looking for three arithmetic progressions whose first terms form an arithmetic progression as well.

I tested all values in the range from 2 up to 10**17. Within this range, no triple of progressions exists that would allow a magic square where all entries satisfy the required conditions. However, I did find three “candidate sets” consisting of two progressions each:

r = 10899799911000

(17162915451926171, 17173815251837171, 17184715051748171)

(17452709490725471, 17463609290636471, 17474509090547471)

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r = 899910000899910

(35480905350908453, 36380815351808363, 37280725352708273)

(37380715051708373, 38280625052608283, 39180535053508193)

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r = 1091001010009110

(15191626062619151, 16282627072628261, 17373628082637371)

(35280837273808253, 36371838283817363, 37462839293826473)

So each of these sets yields at least two magic squares (the base one and the transformed square (m = 2*n + 1)), each of which has at least six numbers satisfying all the conditions. After analysing all possible permutations, I found that using the progressions with r = 899910000899910 one can construct the pair of magic squares shown above. According to my computations, this configuration maximizes the number of entries satisfying the puzzle conditions among all combinations obtainable from these progressions, although there also exists a configuration in which the base square contains 6 numbers satisfying all conditions plus 1 number that is only a palindrome, while in the transformed square there are only 6 numbers satisfying the required conditions.

Therefore, if a pair of magic squares satisfying all conditions exists, my results suggest that the base square must contain values larger than 10**17. 

I am not entirely certain that all computations are correct, but they seem to me to be consistent.

Later on May 14, 2025 she sent:

Hello, in the meantime, I expanded the scope of my search and ultimately examined the range <2; 7*10**18>. The "best" pair I found consists of two squares, each containing exactly 8 numbers that are both prime numbers and palindromes, and exactly 1 term that is only a prime number. I am sending this pair of squares below. Numbers in bold are only prime numbers. The remaining numbers are palindromic primes.

Base magic square:

3636381807081836363    3726371817181736273    3816361827281636183

3906351837381536093    3726371817181736273    3546391796981936453

3636381807081836363    3726371817181736273    3816361827281636183

Transformed square ( m=2*n+1):

7272763614163672727    7452743634363472547    7632723654563272367

7812703674763072187    7452743634363472547    7092783593963872907

7272763614163672727    7452743634363472547    7632723654563272367

 

Unfortunately this last example provided by Arina doesn't follows the golden rule of a magic square "all the integers MUST be DISTINCT", as you can verify in sight.In both squares (not magic) 3636381807081836363 is used twice. The same occurs with 3816361827281636183, An finnaly 3726371817181736273 is repeted thrice. (Qualification 8/18, CR)

One day later, on May 15, 2025, Arine wrote again:

The square that contains 9 unique values and satisfies the most conditions is the following square: 

1925356557556535291    1529480708070849251    1727253646463527271

1529260726270629251    1727363637363637271    1925466548456645291

1727473628263747271    1925246566656425291    1529370717170739251

Transformed square:

3850713115113070583    3058961416141698503    3454507292927054543

3058521452541258503    3454727274727274543    3850933096913290583

3454947256527494543    3850493133312850583    3058741434341478503

The bolded numbers (2) are only palindromes but not primes, and the underlined ones (3) are neither palindromes nor prime numbers. So, 13 integers in both 3x3 squares are Palprimes.

(Qualification 13/18 = 72.22%, Primality & Magicity checked by CR)  

Ultimately, I analyzed the range <2; 10**23> and did not find a better solution containing 9 unique numbers, and I believe that further searching in this way is futile, at least on a home computer due to the enormous computation time required.

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In my opinion, this last solution is a kind of near to the complete palprime solution (13/18=72.22%) but still a kind of far. Perhaps the exact solution could arrive soon... who knows!? CR