Question

Prove that if x and y are both odd positive integers then X square + y square is even but not divisible by 4.​

Answer 1

Given that both x and y are odd positive integers.

As odd numbers leave remainder on division by 2,

let

x = 2a + 1

&

y = 2b + 1.

for any two distinct whole number a and b.

  ⇒ x² + y²

  ⇒ (2a + 1)² + (2b + 1)²

  ⇒ (4a² + 4a + 1) + (4b² + 4b + 1)

  ⇒ 4a² + 4a + 1 + 4b² + 4b + 1

  ⇒ 4a² + 4b² + 4a + 4b + 2

Consider this 4a² + 4b² + 4a + 4b + 2.

  ⇒ 4a² + 4b² + 4a + 4b + 2

  ⇒ 2(2a² + 2b² + 2a + 2b + 1)

Here, as no remainder is obtained, x² + y² is divisible by 2.

  ⇒ 4a² + 4b² + 4a + 4b + 2

  ⇒ 4(a² + b² + a + b) + 2

Here, remainder 2 is obtained on division by 4. Thus x² + y² is not divisible by 4.

Hence proved!!!

Answer 2

Step-by-step explanation:

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Step-by-step explanation:

Given that both x and y are odd positive integers.

As odd numbers leave remainder on division by 2,

let

x = 2a + 1

&

y = 2b + 1.

for any two distinct whole number a and b.

  ⇒ x² + y²

  ⇒ (2a + 1)² + (2b + 1)²

  ⇒ (4a² + 4a + 1) + (4b² + 4b + 1)

  ⇒ 4a² + 4a + 1 + 4b² + 4b + 1

  ⇒ 4a² + 4b² + 4a + 4b + 2

Consider this 4a² + 4b² + 4a + 4b + 2.

  ⇒ 4a² + 4b² + 4a + 4b + 2

  ⇒ 2(2a² + 2b² + 2a + 2b + 1)

Here, as no remainder is obtained, x² + y² is divisible by 2.

  ⇒ 4a² + 4b² + 4a + 4b + 2

  ⇒ 4(a² + b² + a + b) + 2

Here, remainder 2 is obtained on division by 4. Thus x² + y² is not divisible by 4.

Hence proved!!!

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