Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
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Sign up to join this communityFind a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
I believe this is optimal (unless I've missed a trick):
1 13 47 53 29 59 7 19 41 11 37 25 43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59 53 29 23 9 47 19 43 5 13 49 11 41
and...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49 43 41 37 5 59 17 31 19 23 39 11 53
First I found these two:
and1 11 41 61 47 31 17 19 43 13 53 5 23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53 29 59 7 19 61 31 17 5 23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
The following magic square
11 1 53 37 7 47 29 19 71 23 3 5 13 31 17 41
has magic constant
102
and largest number
71