Question

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

Answer 1

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Answer 2

Step-by-step explanation:

Solution: 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.

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