Question

show that square square of any positive integer of the form 3m or 3m+1 where m is an integer​

Answer 1

given:

$let \: a \: be \: any \: positive \: integer \\ \\ using \: euclids \: lemma \: = a = bq + r \\ \\ here \: b = 3 \: then \: possible \: remainders \: are \: = 0.1.2 \\ \\ now \: a \: can \: be \: 3m...3m + 1 ....3m + 2 \\ \ \\ \\ s.o.b.s \\ \\ {a}^{2} = {3m }^{2} \\ {a}^{2} = 9 {m}^{2} \\ {a}^{2} = 3( {3m}^{2} ) \\ \\ {a}^{2} = 3k \: where \: k = 3 {m}^{2}$

${a}^{2} = {3m + 1}^{2} \\ \\ {a}^{2} = {3m}^{2} + 2 \times 3m \times 1 + 1 \\ \\ {a}^{2} = {9m}^{2} + 6m + 1 \\ \\ {a}^{2} = 3( 3{m}^{2} + 2m) + 1 \\ \\ {a }^{2} = 3k + 1 \: \: where \: k \: = 3 {m}^{2} + 2m$

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