Question

Show that the square of any positive integers is of the form 4m and 4m + 1 where m is any positive integers​

Answer 1

SEE UR N.C.E.RT BOOK

PG. NO 6

U GOT THE ANSWER

Answer 2

Hello ,

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Let ,

(q)= m

(${m}^{2} + m$)

X = 2q and X = 2q + 1 is an positive integers

${x}^{2} = 4q \\ {x}^{2} = 4(q) \\ \\ {x}^{2} = 4m$

${x}^{2} = ({2q + 1})^{2} \\ {x}^{2} = \: {2q}^{2} + {1}^{2} + 2 \times 2q \times 1 \\ {x}^{2} = \: {4q}^{2} + 1 + 4q \\ {x}^{2} = \: 4( {q}^{2} + q ) + 1 \\ {x}^{2} = \: 4m + 1$

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