Question

Show that square of any positive integer is either of form 4q or 4q 1 for some integer q

Answer 1

Hey there!

According to Euclid division lemma, There exists two integers q, r for a, b which are positive integers such that ,

$\boxed {a = bq + r, 0 ≤ r < b. }$

So, let's consider a = a, b = 4 .

Now, a = 4x + r.

From the above equation,
0 ≤ r < 4 ,
So, r = { 0 , 1 , 2 , 3 }

When r = 0,
then
a = 4x + 0
Squaring on both sides,
a² = ( 4x)²
a² = 16x²
a² = 4 ( 4x²)
a² = 4q , where q = 4x .

When , r = 1
a = 4x + 1
a² = 16x² + 1 + 8x
a² = 4 ( 4x² + 2x ) + 1
a² = 4q + 1 , where q = 4x² + 2x

When r = 2 ,
a = 4x + 2
a² = 16x² + 4 + 16x
a² = 4 ( 4x² + 4x + 1 )
a² = 4q , where q = 4x² + 4x + 1

When r = 3 ,
a = 4x + 3
a² = 16x² + 9 + 24x
a² = 4( 4x² + 6x + 2 ) + 1
a² = 4q + 1 , where q = 4x² + 6x + 2

Therefore , It is proved that Square of any positive integer is of the form 4q or 4q+ 1 for some integer q.

Answer 2

Step-by-step explanation:

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.