Question

show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q?

Answer 1

hyy friend here is ur answer
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Let positive integer a = 4m + r ,
By division algorithm we know
here 0 ≤ r < 4 , So
When r = 0
a = 4m
Squaring both side , we get
a 2 = ( 4m ) 2
a 2 = 4 ( 4m​ 2 )
a 2 = 4 q , where q = 4m 2
When r = 1
a = 4m + 1
squaring both side , we get
a 2 = ( 4m + 1) 2
a 2 = 16m 2 + 1 + 8m
a 2 = 4 ( 4m 2 + 2m ) + 1
a 2 = 4q + 1 , where q = 4m 2 +
2m
When r = 2
a = 4m + 2
Squaring both hand side , we
get
a 2 = ( 4m + 3) 2
a 2 = 16m 2 + 9 + 24m
a 2 = 16m 2 + 24m + 8 + 1
a 2 = 4 ( 4m 2 + 6m + 2) + 1
a 2 = 4q + 1 , where q = 4m 2 +
6m + 2
Hence
Square of any positive integer
is in form of 4q or 4q + 1 ,
where q is any integer..
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if u like my answer then mark as brainliast
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#ar

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.