Question

show that the square of any odd positive integer 4q+1 or 4q+3 where q is the positive integer.

Answer 1

Hey

Here is your answer,

By Euclid’s division algorithm, we have a = bq + r, where 0
On putting b = 4 in Eq. (i),
we get
a = 4q + r, where 0 ≤ r < 4 i.e., r = 0,1,2,3 ...(ii)

If r = 0 ⇒ a = 4q, 4q is divisible by 2 ⇒ 4q is even .

If r = 1 ⇒ a = 4q + 1,(4q + 1) is not divisible by 2.

If r = 2 ⇒ a = 4q + 2,2(2q + 1) is divisible by 2 ⇒ 2(2q+1) is even.

If r = 3 ⇒ a = 4q + 3, (4q + 3) is not divisible by 2.

So, for any positive integer q, (4q+1) and (4q+3) are odd integers.

Now,
a2 = (4q+1)2 = 16q2 + 1 + 8q = 4(4q2 + 2q) + 1
[∵(a+b)2 = a2 + 2ab + b2] is a square which is of the form 4m + 1, where m = (4q2 + 2q) is an integer.
and a2 = (4q + 3)2 = 16q2 + 9 + 24q = 4(4q2 + 6q + 2) + 1 is a square [∵(a+b)2 = a2 + 2ab + b2]
which is of the form 4m + 1, where m = (4q2 + 6q + 2) is an integer.

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

Hope it helps you!

Answer 2

Step-by-step explanation:

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved .