Question

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

Answer 1

Let a and q be any positive integer and 2 be the divisor. Then by Euclid's division lemma possible outcomes -

a = 2q, a²=(2q)² => 4q² =4m²(where m is any positive integer)
a=2q+1, a²=(2q+1)² => 4q²+ 1 + 2(2q)(1)=> 4q² +1 + 4q => 4(q²+q)+ 1= 4m +1 ( where m is any positive integer, here m= q²+q.
Hence proved

Answer 2

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.

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