Question

use euclid's division algonthem show that cube of any positive integer is of the form 4m,4m+1 or 4m+3 ​

Answer 1

Step by step explanation

Let

n= bq+r (Euclid's division algorithm)

b=4

r=0,1,2,3

Case 1 when r=0

${\sf →n=4q+0 }$

${\sf →n³=(4q)³ }$

${\sf →n³=64q³ }$

${\sf →n³=4(16q³) }$

${\sf →n³=4m ~~~~~~~~~ (16q³=m) }$

Case 2 when r=1

${\sf →n=4q+1 }$

${\sf →n³=(4q+1)³ }$

${\sf →n³=(4q)³+(1)³+3×(4q)²×1+ 3×1²×4q }$

${\sf →n³=64q³+1+3×16q²+ 12q }$

${\sf →n³=64q³+1+48q²+ 12q }$

${\sf →n³=4(16q³+12q²+3q)+1 }$

${\sf →n=4m+1~~~~~~~~~~~~~(16q³+12q²+3q=m) }$

Case 3 when r=2

${\sf →n=4q+2 }$

${\sf →n³=(4q+2)³ }$

${\sf →n³=(4q)³+(2)³+3×(4q)²×2+ 3×2²×4q }$

${\sf →n³=64q³+8+6×16q²+ 48q }$

${\sf →n³=64q³+8+86q²+ 47q }$

${\sf →n³=4(16q³+24q²+12q+2) }$

${\sf →n=4m~~~~~~~~~~~~~(16q³+24q²+12q+2=m) }$

Case 4 When r=3

${\sf →n=4q+2 }$

${\sf →n³=(4q+3)³ }$

${\sf →n³=(4q)³+(3)³+3×(4q)²×3+ 3×3²×4q }$

${\sf →n³=64q³+27+9×16q²+ 12×9q }$

${\sf →n³=64q³+27+144q²+ 108q }$

${\sf →n³=4(16q³+8 +36q²+27q) }$

${\sf → n=4m~~~~~~~~~~ (16q³+8 +36q²+27q=m) }$

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