Question

Prove that the cube of any positive integer is of the form 8m or 8m+1.

Answer 1

Let a be a positive integer.
According to Euclid division lemma,
a = bq + r where 0 ≤ r < b.

Let b = 8 ,
then, a = 8q + r

r can be 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 .

Let's consider r = 0
them a = 8q

Cubing on both sides we get,

a³ = (8q)³
= 512q³
= 8(64q³)
= 8m where m = 64q³

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If r = 1 ,
then a = 8q +1

a³ = (8q + 1)³
a³ = 512q³ + 1 + 3(8q)(8q+1)
= 512q³ + 1 + 24q(8q+1)
= 512q³ + 1 + 192q² + 24q
= 8( 64q³ + 24q² + 3q) + 1
= 8m + 1 where m = 64q³ + 24q² + 3q

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If r = 2 ,

a = 8q + 2
a³ = (8q+2)³
= 512q³ + 8 + 48q(8q+2)
= 512q³ + 8 + 384q²+ 96q
= 512q³ + 384q²+ 96q + 8
= 8 ( 64q³ + 48q² + 12q + 1 )
= 8m
where m = 64q³ + 48q² + 12q+ 1

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if r = 3

then a = 8q + 3
a = (8q+3)³
= 512q³+27+72q(8q+3)
= 8(64q³+ 3 + 72q²+ 27q ) + 3
= 8m+ 3
where m = 64q³ + 72q² + 27q+3)

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If r = 4
a = 8q + 4
a³ = 512q³+ 64 + 768q² + 384q
a³ = 8( 64q³ + 8 + 96q² + 48q)

a³ = 8m
where m = 64q³ + 8 + 96q² + 48q

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when r = 5
a = 8q + 5
a³ = (8q+5)³
= 512q³ + 960q² + 600q + 125
= 8 ( 64q³ + 120q² + 75q + 15) + 5
= 8m + 5

where m = 64q³ + 120q² + 75q + 15
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if r = 6 .

then a = 8q + 6
a³ = ( 8q + 6)^3
= 512q³ + 1152q² + 864q + 216
= 8 ( 64q³ + 144q² + 108q + 27 )
= 8m

where m = 64q³ + 144q² + 108q + 27

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if r = 7
a = 8q + 7
a³ = (8q + 7)³
= 512q³ + 343 + 1344q² + 1176 q
= 8( 64q³ + 168q² + 147q + 42) + 7
= 8m + 7

where m = 64q³ + 168q² + 147q + 42
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Therefore, We proved that cube of a positive integer will be of the form 8m , 8m+ 1 , 8m+3 , 8m+5 , 8m+ 7 where m is a whole number.