Alternate calculation of Julian leap year

[rengel-de]

May I suggest to replace the functions

  def leap_year?(year) when year > 0 do
    Math.mod(year, 4) == 0
  end

  def leap_year?(year) do
    Math.mod(year, 4) == 3
  end

in module Calendrical.Calendar.Julian by an alternative that doesn't use the Math helper module and uses only one function:

def leap_year_alt?(year) do
    (year > 0) && (rem(year, 4) == 0) || (rem(year, 4) == -1)
  end

Afterall we know that year is an integer. The term (rem(year, 4) == -1) takes into account that Erlang produces a negative modulo for negative numbers. In other languages (i.e. Python) this would have been (year % 4) == 3 for negative years.

I tested the new function against the old one using:

  test "Julian.leap_year?(year)" do
    for year <- -4717..2020 do
      assert Julian.leap_year_alt?(year) === Julian.leap_year?(year)
    end
  end

All tests produced the same results and passed.
Of course, if accepted, leap_year_alt?, should be renamed leap_year?, so that no outer depencies will be broken.