1.

Show that any square of any costurent integer for integer q, is the form of 4q or 4q+1

Answer»

ong>Step-by-step explanation:

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

a2 = ( 4m )2

a2 = 4 ( 4M2)

a2 = 4 q , 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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