Question

show that any cube numbers is in the form of 4m,4m+1 or 4m+3
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Answer 1

I hope this is help you.

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Answer 2

To prove :-

Any cube numbers is in the form of 4m,4m+1 or 4m+3

Proof :-

Let's consider 'a' be any positive integer and b = 4.

Then by using Euclid Division Lemma,

a = bq + r [ 0 ≤ r < b ]

a = 3q + r [ 0 ≤ r < 4 ]

So, the values of r be :

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

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If, we put r = 0

a = 4q + 0

a = 4q

On cubing both sides,

▪️ a³ = (4q)³

▪️ a³ = 4 (16q³)

▪️ a³ = 9m [16q³ = m as integer]

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If we put , r = 1

a = 4q + 1

On cubing both sides ;

▪️ 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 [ Take m as some integer ]

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If we put, r = 2,

a = 4q + 2

On cubing both sides :

▪️ 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 [Consider m as some integer]

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If we put , r = 3

a = 4q + 3

On cubing both the sides;

▪️ 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 [Consider m as some integer]

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