Question

Show that square of any odd integer is of the form 4q+1 for some integer q

Answer 1

Step-by-step explanation:

Note :- I am taking q as some integer.

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 ( 4m​²)

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.

THANKS

#BeBrainly.

Answer 2

Step-by-step explanation :

Let any positive odd integer be a. We know that,

a = 2m + 1

On squaring both sides,

a² = ( 2m + 1 )²

a² = 4m² + 4m + 1

a²= 4( m² + m ) + 1

According to Euclid's Lemma, a = 4q + 1 where 0 ≤ r < 4

So, a² = 4q + 1 ( where m² + m = q )

Q.E.D