A
proof of
lisapple's unproven statement:
For every prime > 3 (by the above logic), there exists a number n such that said prime is equal to either b or c, where:
b = 6n+1, c = 6n-1.
Therefore:
b^2 = 36n^2+12n+1
c^2 = 36n^2-12n+1
The difference between any two primes >3, therefore, are equal to:
(b1)^2-(b2)^2 = 36n^2 - 36m^2 +12n - 12m = 12 (3n^2 - 3m^2 + n - m).
The value in the parentheses is even for all integer values of m and n.
Therefore, there exists a number L such that (a1)^2 - (a2)^2 = 24L.
And 24 divides 24L.
The other cases are solvable by similar logic.
Q.E.D.