Question

Prove that square of any +ve integer is of the form 3k or 3k+1

Answer 1

let, any positive integer=p

divide it by 3

then, possible value=3p or 3p+1 or 3p+2[r=0,1,2]

after squaring, we obtain

(3p)^2 or (3p+1)^2 or (3p+2)^2

=9p^2 or 9p^2+6p+1 or 9p^2+12p+4

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

let, 3p^2=k or 3p^2+2p=k or 3p^2+4p=k

then, 3k or 3k+1 or 3k+4

so, square of any integer is of the firm of 3k or 3k+1 or 3k+4 for some positive integer k.

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