What is the pattern of the differences of trios of primes bigger than 3 which are in arithmetic progression? Answer: all the trios have differences divisible by 6.
Alex On won the prize.
Proof: Every prime bigger than 5 must be of the form
because 6n, 6n+2 and 6n+4 are divisible by 2, and 6n+3 is divisible by 3.
In other words, the primes must be equal to 1 or 5 in “clock arithmetic” with six hours, or, as we say in maths, modulo 6.
Say the first prime is 1. Then the second prime in the arithmetic progression must be 1 or 5. So the difference must be 0 in six-hours-clock-arithmetic (i.e. a number divisible by 6), or 4. But if it’s 4, then the third number will be 5+4 = 3 in six-hours-clock-arithmetic, so not a prime. Therefore the difference must be 0.
Similarly, if the first prime is 5, the difference must be 0 or 2 (still in six-hours-clock-arithmetic), but it can’t be 2 because otherwise the third number would be 3, and so not a prime.
Either way, the difference must be 0 in six-hours-clock-arithmetic (i.e. a number divisible by 6). ▇