Question

Show that the square of any odd integer is of the form 4p+1 where p is some integer.

Answer 1

q2 = 4q + 1
q = 2q + 1
q - 2q = 1
-q = 1
q = -1

Answer 2

$\huge\sf{\underline{\underline{Answer}}}$

Here we will use the [ Euclid's division lemma .]

→ a = bq + r

and 0 ≤ b > r

Here, b = 4

→ 0 ≤ 4 > r

•°• r = 0,1,2,3

taking , r = 0

→ a = bq + r = 4q + 0 = 4q

→(4q)² = 16q² = 4(4q²)

Let 4q² be p

→ 4p.

Taking , r = 1

→ a = bq + r ==> 4q + 1

→ (4q + 1)² = (4q)² + 2(4q)(1) + (1)²

→ 16q² + 8q + 1

→ 4(4q² + 2q) + 1

Let 4q² + 2q be p

→ 4p + 1

Taking , r = 2.

→ a = bq + r ==> 4q + 2

→ (4q + 2)² = (4q)² + 2(4q)(2) + (2)²x

→ 16q² + 16q + 4

→ 4(4q² + 4q + 1)

Let 4q² + 4q + 1 be p

→ 4p

Hence, proved

#BAL

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