Is there a 3x3 magic square made up of nine different prime numbers?
arithmetic/prime.magic.square New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Is there a 3x3 magic square made up of nine different prime numbers? Can it be done with primes less than 100? [ Editor: problem statement reconstructed from the reply; the original question is not in the collected material. An earlier poster's example had used the number 1. ]
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"NATHANIEL SILVER" <mathelp@worldnet.att.net> said: >The difference between your example and the one >I supplied is that 1 is only divisible by itself and 1. >Incidentally, do you know why 1 is not prime? Yes, I do. I'm just about ready to put it in the FAQ, from all the eighth-graders who come into Usenet thinking that the primality of 1 is one of the great unsolved philosophical debates of our age. In a group, elements that are factors of the identity are called _units_. An element that is not a unit is called _prime_ if it cannot be factored into two elements that are not units. You would be just as well off asking if 1 is blue as to ask if it's prime. [ Editor: units and primes are defined in a ring (here the integers); in a group every element is a factor of the identity. ] >And can you find such a magic square? Yes and no. LEMMA (in case you didn't know it): In a magic square, the central entry is 1/3 the common sum. Pf: Let X be the common sum. If the magic square is labelled like this: A B C D E F G H I then (A+E+I) + (B+E+H) + (C+E+G) + (D+E+F) = 4X (A+B+C) + (D+E+F) + (G+H+I) = 3X Subtracting these two equations gives 3E=X, the desired result. QED. Now, assume that there is a solution. From the lemma, the nine numbers satisfy the following condition: A+I = B+H = C+G = D+F = 2E There are four solutions to this problem in prime numbers less than 100: 3+71 = 7+67 = 13+61 = 31+43 = 2(37) 3+79 = 11+71 = 23+59 = 29+53 = 2(41) 3+83 = 7+79 = 13+73 = 19+67 = 2(43) 5+89 = 11+83 = 23+71 = 41+53 = 2(47) From any of these four solutions, we would need to find at least four ways in which three of the eight numbers on the left equals three times the central entry. But none satisfy this requirement. Therefore, there are no prime magic squares with all numbers less than 100. However, using the same method as above, here is an all-prime magic square that is slightly larger, with a common sum of 213: 83 29 101 89 71 53 41 113 59 -- Matthew Daly <mwdaly@pobox.com>, May 1998