question_answer

  Let a be odd. = 4 (An integer) + 1 or 3 = 4m + 1 or 4m + 3 where m is an integer Let a be even [ any odd integer divided by 4 will have remainder 1 or 3] a = 2q a3 = 8q3 = 4 (2q3) = 4(m), m = 2q3 is an integer a3 is of the form 4m or 4m + 1 or 4/K + 3 where m is some positive integer.

Answer:

  Show that cube of any positive integer is of the form 4m or 4m + 1 or 4m + 3 where m is a positive integer.