Question

Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some
integer ‘m’

Answer 1

question may be wrong as I know

because the cube of any number for example 2 =8 but it exactly divided 4 and gives reminder 0 ,cube of 3 =27 but it not exactly divisible by 4 and gives reminder 1 .

you can proceed this as many number you can but you will only get remainder 0 and 1 the question is wrong.

hope it helps

please mark it brainliest beacause it is a difficult question .

thank you dear

Answer 2

Question:

Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

Step-by-step explanation:

Let a be the positive integer and b = 4.

Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4. So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.

(4q)3 = 64q3 = 4(16q3) = 4m, where m is some integer.

(4q + 1)3 = 64q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3) + 1 = 4m + 1, where m is some integer.

(4q + 2)3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2) = 4m, where m is some integer.

(4q + 3)3 = 64q3 + 144q2 + 108q + 27 = 4(16q3 + 36q2 + 27q + 6) + 3 = 4m + 3, where m is some integer.

Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.