show that the cube of any positive integer is of the form 4m or 4m+1 or 4m+3 where m is a whole number
Answers
Given : cube of any positive integer is of the form 4m or 4m+1 or 4m+3
To Find : Prove
Solution:
Without losing generality Any number can be represented by
4k , 4k+1 . 4k+ 2 , 4k+3
(4k)³ = 64k³ = 4(16k³) = 4m
(4k + 1)³ = 64k³ + 3.16k².1 + 3.4k.1 + 1
= = 64k³ + 48k² + 12k + 1
= 4(16k³ + 12k² + 3k) + 1
= 4m + 1
(4k + 2)³ = 64k³ + 3.16k².2 + 3.4k.2² + 8
= = 64k³ + 96k² + 48k + 8
= 4(16k³ + 24k² + 12k + 2)
= 4m
(4k + 3)³ = 64k³ + 3.16k².3 + 3.4k.3² + 27
= 64k³ + 144k² + 108k + 27
= 64k³ + 144k² + 108k + 24 + 3
= 4(16k³ + 36k² +27k + 6) +3
= 4m +3
cube of any positive integer is of the form 4m or 4m+1 or 4m+3
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