Question

Show that any positive odd integer of the form 4m + 1 or 4m +3 where m is some positive integer

Answer 1

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.

hope it will help you friend

Answer 2

Answer:

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