Math, asked by yadagirisowmya2006, 19 days ago

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

Answers

Answered by saichavan
12

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}

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