Question

Prove that square of any positive integer is in the form of 4 m and 4 m + 1 where ms same integer

Answer 1

Solution:

Let a be any positive integer and b = 2.

Then, by Euclid's algorithm,

a = bq + r           [0 ≤ r < b]

⇒ a + 2q + r      [0 ≤ r < 2]

Therefore, the possible value of r can be,

• r = 0
• r = 1

Putting the value of r = 0,

⇒ a = 2q + 0

⇒ a = 2q

On squaring both sides ;

⇒ a² = (2q)²

⇒ a² = 4q²

⇒ a² = 4m          [q² = m]

_______________________

Putting the value of r = 1,

⇒ a = 2q + 1

On squaring both sides ;

⇒ a² = (2q + 1)²

⇒ a² = (2q)² + 1² + 2 * 2q * 1

[Since, (a + b)² = a² + b² + 2ab]

⇒ a² = 4q² + 1 + 4q

⇒ a² = 4 (q² + q) + 1

⇒ a² = 4m + 1      [q² + q = m]

_______________________

Hence, the square of any positive integer is in the form of 4m or 4m + 1.

Answer 2

Answer:

See the attachment please for answer!!

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