Pumpe timberuers show that the square of any positive odd integer inof

1.	Pumpe timberuers show that the square of any positive odd integer inof the form 4q4i an and ag for some integer &Set
Answer» Let positive integer a be the any positive integer. Then, b = 4 . By division algorithm we know here 0 ≤ r < 4 , So r = 0, 1, 2, 3. When r = 0 a = 4m Squaring both side , we get a² = ( 4m )² a² = 4 ( 4m²) a² = 4q , where q = 4m² When r = 1 a = 4m + 1 squaring both side , we get a² = ( 4m + 1)² a² = 16m² + 1 + 8m a² = 4 ( 4m² + 2m ) + 1 a² = 4q + 1 , where q = 4m² + 2m When r = 2 a = 4m + 2 Squaring both hand side , we get a² = ( 4m + 2 )² a² = 16m² + 4 + 16m a² = 4 ( 4m² + 4m + 1 ) a² = 4q , Where q = 4m² + 4m + 1 When r = 3 a = 4m + 3 Squaring both hand side , we get a² = ( 4m + 3)² a² = 16m² + 9 + 24m a² = 16m² + 24m + 8 + 1 a² = 4 ( 4m² + 6m + 2) + 1 a² = 4q + 1 , where q = 4m² + 6m + 2 Hence ,Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.

Pumpe timberuers show that the square of any positive odd integer inof the form 4q4i an and ag for some integer &Set

Answer»

Let positive integer a be the any positive integer.

Then, b = 4 .

By division algorithm we know here

0 ≤ r < 4 , So r = 0, 1, 2, 3.

When r = 0

a = 4m

Squaring both side , we get

a² = ( 4m )²

a² = 4 ( 4m²)

a² = 4q , where q = 4m²

When r = 1

a = 4m + 1

squaring both side , we get

a² = ( 4m + 1)²

a² = 16m² + 1 + 8m

a² = 4 ( 4m² + 2m ) + 1

a² = 4q + 1 , where q = 4m² + 2m

When r = 2

a = 4m + 2

Squaring both hand side , we get

a² = ( 4m + 2 )²

a² = 16m² + 4 + 16m

a² = 4 ( 4m² + 4m + 1 )

a² = 4q , Where q = 4m² + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand side , we get

a² = ( 4m + 3)²

a² = 16m² + 9 + 24m

a² = 16m² + 24m + 8 + 1

a² = 4 ( 4m² + 6m + 2) + 1

a² = 4q + 1 , where q = 4m² + 6m + 2

Hence ,Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.