We start with the numbers a, b, c, d. We then replace them with a' = a - b, b' = b - c, c' = c - d, d' = d - a. We carry out this process 1996 times. Is it possible to end up with numbers A, B, C, D such that |BC - AD|, |AC - BD|, |AB - CD| are all primes?
Solution
Answer: no.
The first 4 iterations give:
a - b b - c c - d d - a
a-2b+c b-2c+d c-2d+a nbsp; d-2a+b
a-3b+3c-d b-3c+3d-a c-3d+3a-b d-3a+3b-c
2a-4b+6c-4d 2b-4c+6d-4a 2c-4d+6a-4b 2d-4a+6b-4c
So after 4 iterations the numbers are all even. Any common factor is preserved by an iteration, so after 1996 iterations the numbers are all multiples of 2. Hence |BC - AD|, |AC - BD|, |AB - CD| are all multiples of 4, so none of them are prime.
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2002