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

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Let a be any positive integer and b = 4.

According to Euclid Division Lemma,

a = bq + r [0 ≤ r < b]

a = 3q + r [0 ≤ r < 4]

According to the question, the possible values of r are,

r = 0, r = 1, r = 2, r = 3

When r = 0,

a = 4q + 0

a = 4q

Taking cubes on LHS and RHS,

We have,

a³ = (4q)³

a³ = 4 (16q³)

a³ = 9m         [where m is an integer = 16q³]

When r = 1,

a = 4q + 1

Taking cubes on LHS and RHS,

We have,

a³ = (4q + 1)³

a³ = 64q³ + 1³ + 3 × 4q × 1 (4q + 1)

a³ = 64q³ + 1 + 48q² + 12q

a³ = 4 (16q³ + 12q² + 3q) + 1

a³ = 4m + 1        [where m is an integer = 16q³ + 12q² + 3q]

When r = 2,

a = 4q + 2

Taking cubes on LHS and RHS,

We have,

a³ = (4q + 2)³

a³ = 64q³ + 2³ + 3 × 4q × 2 (4q + 2)

a³ = 64q³ + 8 + 96q² + 48q

a³ = 4 (16q³ + 2 + 24q² + 12q)

a³ = 4m   [where m is an integer =16q³ + 2 + 24q² + 12q]

When r = 3,

a = 4q + 3

Taking cubes on LHS and RHS,

We have,

a³ = (4q + 3)³

a³ = 64q³ + 27 + 3 × 4q × 3 (4q + 3)

a³ = 64q³ + 24 + 3 + 144q² + 108q

a³ = 4 (16q³ + 36q² + 27q + 6) + 3

a³ = 4m + 3 [where m is an integer =16q³ + 36q² + 27q + 6]

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

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