Question

Using euclids division lemma show that square of any positive integer is either of form 3m or 3m + 1

Answer 1

Answer:

Let a be positive integer In the form 3q,3q+1

,3q+2.

case \: 1 =  >  \\ when \: a = 3q \\ so \:  \:  {a}^{2}  =  ( {3q})^{2}   \\  =  > 9  {q}^{2}   \\  =  > 3q(3q) =  > 3m \\ where \: m = 3q

Case 2

when \: a = 3q \\ so \:   \\  =  > {a}^{2} ({3q + 1})^{2}  \\  =  >  9 {q }^{2}  + 6q+ 1 \\  =  > 3 q+  (3q + 2) + 1 \\  

=>  

=  > 3m \:  + 1 \\ where \:  \: m \:  = q(3q + 1)

Case 3

=  >  {a}^{2}  =   \: ( {3q + 2)}^{2}    \\ =  > 9 {q}^{2}  + 12q \:  + 4 \\  =  > 9 {q}^{2}  + 12q + 3 + 1 \\  =  > 3(3 {q}^{2}  + 4q + 1) + 1 \\  =  > 3m + 1 \\ where \: m \:  = 3 {q}^{2}  + 4q + 1 \\  

Hence ,a is of the form 3m, 3m+1

Answer 2

let be a positive integer and B=3 euclud lemma a=3q+r
where 0=<r=<3
r=0,1,2
r=0
a=3q
squaring on both side
${a}^{2} = {3q}^{2} \\ {a}^{2} = {9q}^{2} \\ {a}^{2} = 3 \times {3q}^{2} \\ {a}^{2} = 3 \\ where \: m \: = {3q}^{2} isam \\ n \: integer$
other are similar
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