Math, asked by niranjan0, 1 year ago

Show that the square of any odd positive integer is in the form (8m+1), for some integer m.

Answers

Answered by BloomingBud
10

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.


kashifashareef: Very satisfied answer. thanks a lot!!!!
BloomingBud: :)
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