Show that square of odd positive integer is of the Form 8m + 1 for som

1.	Show that square of odd positive integer is of the Form 8m + 1 for some whole number m
Answer» $\huge \mathfrak \red{answer}$ _____________________________ 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 ____________________________ $\sf \underline \blue{case1}$ $= \rm{r = 1}$ $= \rm{a = bm + r}$ $= \rm{(8q + 1) }^{2}$ $= \rm{64 {m}^{2} + 16m + 1}$ $\rm{= 8( 8 {m}^{2} + 2m ) + 1}$ $= \rm{8q + 1 . [ Where \: q = 8 {m}^{2} + 2m ]}$ ___________________________ $\sf \underline \blue{case2}$ $= \rm{r = 3}$ $\rm{a = bq + r}$ $\rm{= (8q +3) }^{3}$ $\rm{= 64 {m}^{2} + 48m + 9 = 64 {m}^{2} + 48m + 8 + 1}$ $\rm{= 8( 8 {m}^{2} + 6m + 1 ) + 1}$ $= \rm{8q + 1 . [ Where \: q = 8 {m}^{2} + 6m + 1 ]}$ _________________________ $\sf \underline \blue{case3}$ $\rm{r = 5}$ $\rm{a = (8q + 5) ^{2} }$ $\rm{= 64 {m}^{2} + 80m + 25 = 64 {m}^{2} + 80m + 24 + 1}$ $\rm{= 8( 8 {m}^{2} + 10m + 3 ) + 1}$ $\rm{= 8q + 1 . [ Where \: q = 8 {m}^{2} + 10m + 3}$ ____________________________ $\sf \underline \blue{case4}$ $\rm{r = 7}$ $\rm{a = ( 8q + 7)^{2} }$ $\rm{= 64 {m}^{2} + 112m + 49 = 64 {m}^{2} + 112m + 48 + 1}$ $\rm{= 8( 8 {m}^{2} + 14m + 6 ) + 1}$ $\rm{= 8q + 1 . [ Where q = 8 {m}^{2} + 14m + 6 ] }$ therefore,the square of any odd positive integer is of the form 8q + 1 . I hope it's help uu mark it as brainlist