Question

8. Use Euclid's division lemma to show that the square of any positive integer is either of the form 4 k or 4 k + 1 for some integer k. Hence prove that 530 cannot be a perfect square.

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Answer 1

Hey Mate !

Here is your solution :

First,we need to know what is Euclid's Division Lemma ?

Euclid's Division Lemma states that for two positive numbers a and b such that a > b , 2 numbers q and r exists such that ,

=> a = bq + r

Where , 0 ≤ r < b.

Now,

Let any number is a and it is divided by 2.

So,

By Euclid's Division Lemma ,

=> a = 2q + r ( 0 ≤ r < 2 )

So,

Possible values of r are 0 and 1.

In case 1,

When , r = 0.

=> a = 2q

Let q = m.

So, a = 2m

In case 2,

When , r = 1.

=> a = 2q + 1

Let , q = m.

So, a = 2m + 1

In case 3,

When , r = 2.

=> a = 2q + 2

=> a = 2 ( q + 1 )

Let , ( q + 1 ) = m.

So, a = 2m

Hence, any positive number can be in either form of 2m or 2m + 1.

Now,

In case 1,

=> a = 2m

By squaring both sides,

=> a² = ( 2m )²

=> a² = 4m²

=> a² = 4 ( m² )

Suppose , m² = k.

=> a² = 4k

Case 2,

When,

=> a = 2m + 1

Squaring both sides,

=> a² = ( 2m + 1 )²

=> a² = ( 2m )² + 1² + 2 × 2m × 1

=> a² = 4m² + 1 + 4m

=> a² = 4m² + 4m + 1

=> a² = 4 ( m² + m ) + 1

Let, ( m² + m ) = k,

=> a² = 4k + 1.

Hence, the square of any positive number is in the form of either 4k or 4k + 1.

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Let, a = 530 and b = 4.

Using Euclid's Division Lemma ,

=> 530 = 4 × 132 + 2

Let,

132 = k.

So,

=> 530 = 4k + 2

We have just seen that a positive number is a perfect square only and only if it is in the form of either 4k or 4k + 1 , but 530 is neither in the form of 4k nor ( 4k + 1 ).Hence, it is not a perfect square

Proved !!

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Hope it helps !! ^_^

Answer 2

Hey!!!

As promised I am here to help you

Good Morning

Difficulty Level : Above Average

Chances of being asked in Board : 75%

This is the official NCERT method so you can trust it

________________

let n be any positive integer

let b = 2(Divisor)

Thus by Euclid's Division Lemma,

=> n = 2q + r where 0 < = r < b

Thus r = 0 or 1

Case 1 when r = 0

=> n = 2q

Square both sides

=> n² = 4q²

=> n² = 4k where k = q²

Case 2 when r = 1

=> n = 2q + 1

Square both sides

=> n² = (2q + 1)²

=> n² = 4q² + 1 + 4q

=> n² = 4k + 1 where k = q² + q

Thus from Case 1 and Case 2

We can say square is any positive integer is of the form 4k or 4k + 1

HENCE PROVED

Part 2

=> 530 = 4(132) + 2

Here it is not of the form 4k or 4k + 1

Hence 530 cannot be a perfect square

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Hope this helps ✌️