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

By Euclid’s division algorithm
a = bq + r, where 0 ≤ r ≤ b
Put b = 4
a = 4q + r, where 0 ≤ r ≤ 4
If r = 0, then a = 4q even
If r = 1, then a = 4q + 1 odd

If r = 2, then a = 4q + 2 even
If r = 3, then a = 4q + 3 odd
Now, (4q + 1)² = (4q)² + 2(4q)(1) + (1)²
= 16q² + 8q + 1
= 8(2q² + q) + 1
= 8m + 1 where m is some integer
Hence the square of an odd integer is of the form 8q + 1, for some integer q

To show:

the square of an odd positive integer is of the form 8q + 1, for some integer q.

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.

When n = 4p + 1,

then n² = (4p + 1)²

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.

Note:

To show that the square of an odd positive integer is of the form 8q + 1

We have started the question from taking b = 4 initially because when we take square of any form of 4 such as 4p+1 we end up having the values which are the multiple of 8 as in (1).

While attempting these types of questions always remember to start with the value which would end up giving the value the question demands.and follow the above steps.