Question

Nice question.
Give it a try for this challenge.

Prove that the remainder of any perfect square divided by 4 is always between 0 or 1.
Further, decide the remainder of perfect squares divided by 3 and prove that the remainders are recurring.

Hint:
$\begin{cases}2k-1 & \text{(odd)}\\2k & \text{(even)}\end{cases}$

Answer 1

Solution :

{ a }

Let us assume that ∃ a perfect square , k = s² .

Now, s can be both even or odd .

Let us take both cases .

When s € Even ,

Let s = 2a , ∀ a € N .

k = s² = ( 2a )² = 4a²

4a² ≡ 0 ( mod 4 )

When k is an even perfect square, the remainder when divided by 4 is 0 .

When s € Odd

Let s = 2b + 1 , ∀ b € N .

k = s² = ( 2b + 1)² = 4b² + 4b + 1 .

4b² + 4b + 1 ≡ 1 ( mod 4 )

When k is an odd perfect square, the remainder when divided by 4 is 1 .

Hence Proved .

{ b }

When any number is divided by 3 , the residual classes are 0, 1 and 2 only . [ Proof of this lemma can be done by induction ] .

That is , all numbers are of the form , 3k , 3k + 1 and 3k + 2 only .

Now ,

Let us assume that ∃ a perfect square , k = l² .

For l = 3k

k = l² = ( 3k )² = 9k²

9k² ≡ 0 ( mod 3 )

For l = (3k + 1 )

k = l² = ( 3k + 1)² = 9k² + 9k + 1 = 3( 3k² + 3k ) + 1

Let 3k² + 3k = m .

k = 3m + 1 , ≡ 1 ( mod 3 )

For l = (3k + 2)

k = l² = ( 3k + 2 )² = 9k² + 12k + 4

=> 9k² + 12k + 3 + 1

=> 3 ( 3k² + 4k + 1 ) + 1

let 3k² + 4k + 1 = n

=> 3n + 1 ≡ 1 ( mod 3 )

The remainder when any perfect square is divided by 3 is 0 or 1 only .

Hence Proved

Answer 2

Answer:

Take common 3 from the possible terms and pair them

→ (3n + 2) = 3(3n2 + 4n + 1) +1

Compare the equation with the general equation

of division i.e. a = bq+r

Remainder = 1

From the above three cases, the remainder is either 0 or 1.

.Remainder of any perfect square when divided

by 3 is either 0 or 1