Question

show that square of odd positive integer is of the Form 8m + 1 for some whole number m ​

Answer 1

$\huge \mathfrak \red{answer}$

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Question:✥

show that square of odd positive integer is of the Form 8m + 1 for some whole number m

step to step explanation:❖

• let 'a' is any positive integer
• then b=8

using euclid division lemma

• 0 ≤ r < b then
• 0 ≤ r < 8 .

then positive eithere are 0,1,2,3,4,5,6,7

from the question given odd positive integer

so it's will be 1,3,5,7

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$\sf \underline \blue{case1}$

$= \rm{r = 1}$

$= \rm{a = bm + r}$

$= \rm{(8q + 1) }^{2}$

$= \rm{64 {m}^{2} + 16m + 1}$

$\rm{= 8( 8 {m}^{2} + 2m ) + 1}$

$= \rm{8q + 1 . [ Where \: q = 8 {m}^{2} + 2m ]}$

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$\sf \underline \blue{case2}$

$= \rm{r = 3}$

$\rm{a = bq + r}$

$\rm{= (8q +3) }^{3}$

$\rm{= 64 {m}^{2} + 48m + 9 = 64 {m}^{2} + 48m + 8 + 1}$

$\rm{= 8( 8 {m}^{2} + 6m + 1 ) + 1}$

$= \rm{8q + 1 . [ Where \: q = 8 {m}^{2} + 6m + 1 ]}$

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$\sf \underline \blue{case3}$

$\rm{r = 5}$

$\rm{a = (8q + 5) ^{2} }$

$\rm{= 64 {m}^{2} + 80m + 25 = 64 {m}^{2} + 80m + 24 + 1}$

$\rm{= 8( 8 {m}^{2} + 10m + 3 ) + 1}$

$\rm{= 8q + 1 . [ Where \: q = 8 {m}^{2} + 10m + 3}$

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$\sf \underline \blue{case4}$

$\rm{r = 7}$

$\rm{a = ( 8q + 7)^{2} }$

$\rm{= 64 {m}^{2} + 112m + 49 = 64 {m}^{2} + 112m + 48 + 1}$

$\rm{= 8( 8 {m}^{2} + 14m + 6 ) + 1}$

$\rm{= 8q + 1 . [ Where q = 8 {m}^{2} + 14m + 6 ] }$

therefore,the square of any odd positive integer is of the form 8q + 1 .

I hope it's help uu

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Answer 2

Answer:

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