Solution To Friday Puzzle #6

9 10 2011

Let black squares have value 1 and white squares have value -1. The entire chessboard has now turned into a base -1 multiplication table of its first row and first column. For example, if a white first row square (having value -1) contains a pawn, and another first column square that is black (having value 1) contains a pawn, then the square on the intersectiong row and column will have value 1*-1 = -1 and hence will be white. Try it with other color combinations as well!

Creating the diagonally symmetric square pattern described in the puzzle is therefore equivalent to computing the square – the 2nd power – of the number designated by the first row. If the first row has 2 blacks and 4 whites then its value is 2*1 + 4*-1 = -2 , the square of which is 4 .  Effecticely this means that four more black squares (+1) are covered than white square (-1) by the complete pattern. Note that the 2nd power can never be negative which means that there can never be more white squares covered than black squares.

So the question of the smallest board containing a difference that is a prime numer is equivalent to asking for the smallest 2nd power of an integer that is a prime, which obviously does not exist. Hence my intended answer was that there is no solution.

However, one reader pointed out to me offline that a square with 1 pawn in (0,0) should qualify as an answer. Intriguing! Some googling quickly revealed that there have been different definitions historically of what is a prime number. The best explanation is provided by Wikipedia:

Most early Greeks did not not even consider 1 to be a number, so clearly did not consider it a prime. In the 19th century however, many mathematicians did consider the number 1 a prime. For example, Derrick Norman Lehmer’s list of primes up to 10,006,721, reprinted as late as 1956, started with 1 as its first prime. Henri Lebesgue is said to be the last professional mathematician to call 1 prime. Although a large body of mathematical work is also valid when calling 1 a prime, the fundamental theorem of arithmetic does not hold as stated. For example, the number 15 can be factored as 3 · 5 or 1 · 3 · 5. If 1 were admitted as a prime, these two presentations would be considered different factorizations of 15 into prime numbers, so the statement of that theorem would have to be modified.

For the reason stated in the last sentence of the above quotation (i.e. preserving unique factoring) the contemporary definition of a prime number explicitly excludes 1 from the set of primes. But because I failed to specify the definition of primes relevant to this puzzle, I feel compelled to accept her solution as a correct one.




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