Question

Show that any positive integer is of form 4 or 4q+1 or 4q+2 or4 +3 where q is some integer.​

Answer 1

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 = 4q , 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 + 2 )^2

a^2 = 16m^2 + 4 + 16m

a^2 = 4 ( 4m^2 + 4m + 1 )

a^2 = 4q , Where q = 4m^2 + 4m + 1

When r = 3

a = 4m + 3

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.

Answer 2

Answer:

Step-by-step explanation:

let a be any positive integer  

by EDL  a = bq +r  

0 ≤ r < b

possible remainders are 0, 1, 2 , 3  

this shows that a can be in the form of 4q, 4q+1, 4q+2, 4q+3 q is quotient  

as a is odd a  can't be the form of 4q or 4q+2 as they are even

so a ill be in the form of 4q + 1 or 4q+3  

hence proved