Question

use division algorithm to show that the square of any positive integer is of the form 3p or 3p +1​

Answer 1

Answer:

$\sf \: Any \: number \: can \: be \: represented \: as \: 3q \: , \: 3q + 1 \: , 3q + 2 \: where \: q \: is \: an \: positive \: integer.$

$\sf \red{ \therefore \: (3 {q}^{2} )}$

$\sf \red{ = 9 {q}^{2} }$

$\sf \red{ \implies \: 3 \times 3 {q}^{2} }$

$\sf \green{ \implies \: 3p}$

$\sf \: But \: q \: is \: an \: integer \implies \: 3q² \: is \: an \: integer$

$\sf \red{ \implies \: (3q + 1) {}^{2} }$

$\sf \red{ \implies(9 {q}^{2} + 6q + 1)}$

$\sf \red{ \implies \:3(3 {q}^{2} + 2q) + 1 }$

$\sf \: But \: 3q² + 2q \: is \: an \: integer \: because \: q \: is \: an \: integer$

$\sf \green{ \therefore = \: 3 {p}^{2} + 1}$

Now ,

$\sf \red{(3q + 2) {}^{2} }$

$\sf \red{ \implies \: 9 {q}^{2} + 12q + 4 }$

$\sf \red{ \implies \: 9 {q}^{2} + 12q + 3 + 1 }$

$\sf \red{ \implies3 ( \: 3 {q}^{2} + 4q + 1) + 1 }$

$\sf \red{3 {p}^{2} + 4q + 1..(is \: integer \because \: q \: is \: an \: integer}$

$\sf \green{ \therefore \: 3p + 1}$

$\sf \green{ Hence \: proved}$