Question

Show that square of any positive integer is of the form 4m or 4m +1, where m is any positive integer.

Answer 1

Answer:
Let a be any positive integer. We apply the division lemma with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1 and 2.
That is, a can be 4q, or 4q + 1, 4q + 2 or 4q + 3 where q is the quotient.

Now, (4q)2 = 16q²
which can be written in the form 4m, since 16 is divisible by 4.

Again,
(4q + 1)² = 16q² + 8q + 1
= 4 (4q² + 2q) + 1
which can be written in the form 4m + 1 since 16q2 + 8q, i.e., 4 (4q² + 2q) is divisible by 4.

Now,
(4q + 2)² = 16q² + 16q + 4
= 4(4q² + 4q + 1)

which can be written in the form 4m , since16q² + 16q + 4, i.e., 4(4q² + 4q + 1) is divisible by 4.

Lastly
( 4q + 3 )² = ( 16q² + 24q + 9 )

= ( 16q² + 24q + 8 ) + 1

= 4 ( 4q² + 6q + 2 ) + 1
which can be written in the form 4m + 1 since 16q² + 24q + 8​, i.e., 4 ( 4q² + 6q + 2 ) is divisible by 4.
Therefore, the square of any positive integer is either of the form 4m or 4m + 1 for some integer m. ( Hence proved )