Question

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

Answer 1

Answer:

hey dear .....

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.

$<marquee> hope it's helpful for you..</marquee>$

Answer 2

Answer:

see attachment for ur question

therefore, cube of any +ve interger is of the form 4m, 4m+1 or 4m+3 for some interger m