Use Euclid's division lemma to show that hte square of any integer is either of the form 5m, 5m+1 or 5m+4 for some integer m.
Answers
Answer:
Step-by-step explanation:
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.