Show that square of an odd digit positive integer is of the form 8m+1,

1.	Show that square of an odd digit positive integer is of the form 8m+1,for some integer
Answer» Let a be the odd positive intezer and b=8.By Euclid division lemma where r=0,1,2,3,4,5,6,7.When r=0a=8q+0When r=1a=8q+1 (odd)When r=2a=8q+2When r=3a=8q+3 (odd)When r=4a=8q+4When r=5a=8q+5 (odd)When r=6a=8q+6When r=7a=8q+7 (odd) According to Euclid division lemma , a = bq + r where 0 ≤ r < bHere we assume b = 8 and r ∈ [1, 7 ] means r = 1, 2, 3, .....7Then, a = 8q + rCase 1 :- when r = 1 , a = 8q + 1squaring both sides,a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1= 8m + 1 , where m = 8q² + 2qcase 2 :- when r = 2 , a = 8q + 2squaring both sides,a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1 [ means when r is an even number it is not in the form of 8m + 1 ]Case 3 :- when r = 3 , a = 8q + 3squaring both sides,a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1= 8m + 1 , where m = 8q² + 6q + 1You can see that at every odd values of r square of a is in the form of 8m +1But at every even Values of r square of a isn\'t in the form of 8m +1 .Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbersHence , it is clear that square of an odd positive is in form of 8m +1.\xa0

Show that square of an odd digit positive integer is of the form 8m+1,for some integer