Question

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

Answer 1

Let 'a' be any positive integer.
b=4
by Euclid's division lemma,
a=bq+r
a²=(bq+r)²-------1.
r=0,1,2,3
from 1.
for r=0,
a²=(4q+0)²
a²=16q²
a²=4(4q²)
=4q, where q=4q².
for r=1,
a²=(4q+1)²
a²=16q²+1+8q
a²=4(4q²+2q)+1
a²=4q+1, where q=4q²+2q.

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.

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