Question

Show that square of any positive integer is of the form 4m or 4m+11 where m is any integer .

Answer 1

Euclid division lemma :-
a=bm+r

b=4,
a=4m+r , 0≤r<4

The possible values of r are 0,1,2 and 3.
Let a be any positive integer.
a=1,2,3,4,5....
Square of any positive integer,
(1)²=1=4(0)+1 [r=1}
(2)²=4=4(1)+0 [r=0]
(3)²=9=4(2)+1 [r=1]
(4)²=16=4(4)+0 [r=0]
(5)²=25=4(6)+1 [r=1]

So the square of any positive integer is of the form 4m or 4m+1 for any positive integer m.
Hence proved.

Hope it helps..

Answer 2

Step-by-step explanation:

Note :- Your question have some error, it will be 4m + 1 . And 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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