Question

युक्लिड विभाजन का कर
चनात्मकन्यीक कार्यक्रम पूर्णाक का वर्गीक
के लिए
Sm, Sam+ 5m +4-
का
रूप होगा ​

Answer 1

$\huge\boxed{\fcolorbox{yellow}{red}{☞ANSWER:-}}$

Let 'a' be any positive integer and b = 5.

Using Euclid Division Lemma,

a = bq + r            [ 0 ≤ r < b ]

⇒ a = 5q + r       [ 0 ≤ r < 5 ]

Now, possible value of r :

r = 0, r = 1, r = 2, r = 3, r = 4

CASE I :-

If we take, r = 0

⇒ a = 5q + 0

⇒ a = 5q

On squaring both sides;

⇒ a² = (5q)²

⇒ a² = 25q²

⇒ a² = 5 ( 5q² )

⇒ a² = 5m.    [ Here, m = 5q² ]

CASE II :-

If we take, r = 1

⇒ a = 5q + 1

On squaring both sides;

⇒ a² = (5q + 1)²

⇒ a² = (5q)² + 2 * 5q * 1 + 1²

⇒ a² = 25q² + 10q + 1

⇒ a² = 5 ( 5q² + 2q ) + 1

⇒ a² = 5m + 1   [ Here, m = 5q² + 2q ]

CASE III :-

If we take, r = 2

⇒ a = 5q + 2

On squaring both sides;

⇒ a² = (5q + 2)²

⇒ a² = (5q)² + 2 * 5q * 2 + 2²

⇒ a² = 25q² + 20q + 4

⇒ a² = 5 ( 5q² + 4q ) + 4

⇒ a² = 5m + 4  [ Here, m = 5q² + 4q ]

CASE IV :-

If we take, r = 3

⇒ a = 5q + 3

On squaring both sides;

⇒ a² = (5q + 3)²

⇒ a² = (5q)² + 2 * 5q * 3 + 3²

⇒ a² = 25q² + 30q + 5 + 4

⇒ a² = 5 ( 5q² + 6q + 1) + 4

⇒ a² = 5m + 4  [ Here, m = 5q² + 6q + 1 ]

CASE V :-

If we take, r = 4

⇒ a = 5q + 4

On squaring both sides;

⇒ a² = (5q + 4)²

⇒ a² = (5q)² + 2 * 5q * 4 + 4²

⇒ a² = 25q² + 40q + 15 + 1

⇒ a² = 5 ( 5q² + 8q + 3 ) + 1

⇒ a² = 5m + 1  [ Here, m = 5q² + 8q + 3 ]

Hence, the square of any integer is either of the form 5m, 5m+1 or 5m+4 for some integer m.

                 Hence  Proved.