==> analysis/period.s <==
Period 2.  Clearly, the sum of periodic functions of periods 2 and
three is 6.  So take the function which is the sum of that function of
period six and the negative of the function of period three and you
have a function of period 2.

This proves that a period-2 solution exists, but not that it is minimal.
Since we're talking about integers, the only lower possibility is 1.
But the sum or difference of a period-1 and a period-3 function must have
period 3, not 6, therefore 1 is indeed impossible.