Show that the square of any odd positive integer is in the form (8m+1), for some integer m.
Answers
SOLUTION :
Let a be any positive integer and b = 8
so,
By applying EDL we can write,
a = 8q + r [ 0 ≤ r < 8 ]
So,
remainder possible values - 0, 1, 2, 3, 4, 5, 6, 7
Odd remainders - 1, 3, 5, 7
CASE 1
when r = 1
a = 8q + 1
(a)² = (8q +1)²
a² = (8q)² + (1)² + 2 × 8q × 1 [(a+b)² = a² + b² +2ab]
a² = 64q² + 1 + 16q
a² = 64q² + 16q + 1
a² = 8 (8q² + 2q) + 1
a² = 8m + 1 ____(i) [taking m = (8q² + 2q)]
CASE 2
when r = 3
a = 8q + 3
(a)² = (8q +3)²
a² = (8q)² + (3)² + 2 × 8q × 3 [(a+b)² = a² + b² +2ab]
a² = 64q² + 9 + 48q
a² = 64q² + 48q + 8 + 1
a² = 8 (8q² + 6q + 1) + 1
a² = 8m + 1 ____(ii) [taking m = (8q² + 6q + 1)]
CASE 3
when r = 5
a = 8q + 5
(a)² = (8q +5)²
a² = (8q)² + (5)² + 2 × 8q × 5 [(a+b)² = a² + b² +2ab]
a² = 64q² + 25 + 80q
a² = 64q² + 80q + 24 + 1
a² = 8 (8q² + 10q + 3) + 1
a² = 8m + 1 ____(iii) [taking m = (8q² + 10q + 3)]
CASE 4
when r = 7
a = 8q + 7
(a)² = (8q +7)²
a² = (8q)² + (7)² + 2 × 8q × 7 [(a+b)² = a² + b² +2ab]
a² = 64q² + 49 + 1112q
a² = 64q² + 112q + 48 + 1
a² = 8 (8q² + 14q + 6) + 1
a² = 8m + 1 ____(iv) [taking m = (8q² + 14q + 6)]
Therefore,
we conclude from the above equation (i), (ii), (iii), (iv) and we can say that
The square of any odd positive integer is in the form (8m+1) for some integer m.