Question

use eculids division lemma to show that the square of any positive integers is the form 3p,3p+1.​

Answer 1

Let a be the positive integer and b=3

therefore, a=bq+r 0< also = r < b

remainder r =0,1,2

therefore when r=0,

a=3p+0=(3p)^2=9p^2=3(3p^2)

= 3p where p= 3p^2

when r=1

a=3p+1 =(3p+1)^2= (3p)^2 + 2(3p)(1)+1^2

=9p^2+6p+1 (3 is common )

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=3(3p^2+2p)+1 = 3m+1 where m= (3p^2+2p)

when r=2

a=3p+2=(3p+2)^2 = (3p)^2+2(3p)(2)+(2)^2

= 9p^2+12p+4 (3 is common)

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=3(3p^2+4+3)+1=3p + 1

4 is spilting to 3+1