Question

Show that square of any positive odd integer is of form 8q+1 for some integer q.

Answer 1

Hi mate here is the answer'-✍️

Solution:

Let a be any positive integer and b=4.

Applying the Euclid's division lemma with a and b=4 we have

a = 4p + r

where 0≤r<4 and p is some integer,

⇒ r can be 0,1,2,3

⇒ a = 4p + 0 ,

a = 4p + 1 ,

a = 4p+ 2 ,

a = 4p + 3

Since a is odd integer,

So a = 4p + 1 or a = 4p + 3 So any odd integer is of the form a = 4p+ 1 or a = 4p+ 3 .

Since any odd positive integer n is of the form 4p + 1 or 4p + 3.

Since any odd positive integer n is of the form 4p + 1 or 4p + 3.When n = 4p + 1,

2 Apply the formula (a + b)2 = a2 + b2 + 2ab

⇒ (4p + 3)2

= 16p2 + 24p + 9

Now 16p2 + 24p + 9 can be written as 16p2 + 24p + 8 + 1

Apply the formula

(a + b)2 = a2 + b2 + 2ab

⇒ (4p + 1)2 = 16p2 + 8p + 1 ..... (1)

Take 8p common out of 16p2 + 8p

⇒ 8p(2p+ 1) + 1 = 8q + 1

where q = p(2p + 1)

If n = 4p + 3,

then n2 = (4p + 3)2

Apply the formula

(a + b)2 = a2 + b2 + 2ab

⇒ (4p + 3)2= 16p2 + 24p + 9

Now 16p2 + 24p + 9 can be written as 16p2 + 24p + 8 + 1

⇒ (4p + 3)2

= 8(2p2 + 3p + 1) + 1

= 8q + 1 where q = 2p2 + 3p + 1

From above results we got that n2 is of the form 8q + 1.

Answer 2

By Euclid's division lemma, given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

Take b = 4, a = 4q + r.

(i) If r = 0:

a = 4q (Even)

(ii) If r = 1:

a = 4q + 1 (Odd)

Now,

(4q + 1)^2

= (4q)^2 + 1 + 2(4q)(1)

= 16q^2 + 1 + 8q

= 8(2q^2 + q) + 1

= 8m + 1, where m is some integer.

Hope it helps!