Question

use Euclid division lemma to show that the square of any positive integer is of the form 3p,3p,3p+1

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Answer 1

Answer:

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Step-by-step explanation:

let us take, 'x'= 3q , 3q+1, 3q+2

when, x=3q

       x2 =  (3q) 2

        x2 = 9q2  

       x2  = 3(3q2)

we see that 3q2= m

so we have done the first equation 3m

when , x=3q+1

          x2= (3q+1)2

                                [since, (a+b)2 = a2+2ab+b2]

          x2= 9q+6q+1

          x2= 3(3q+2q)+1

in this we see that 3q+2q= m

   therefore, this satisfy the equation m+1

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Anshita , Student

Member since Feb 04 2015

let us take, 'x'= 3q , 3q+1, 3q+2

when, x=3q

       x2 =  (3q) 2

        x2 = 9q2  

       x2  = 3(3q2)

we see that 3q2= m

so we have done the first equation 3m

when , x=3q+1

          x2= (3q+1)2

                                [since, (a+b)2 = a2+2ab+b2]

          x2= 9q+6q+1

          x2= 3(3q+2q)+1

in this we see that 3q+2q= m

   therefore, this satisfy the equation m+1