In this problem, we show a property of odd positive integers. We are required to prove that the square of any odd positive integer can always be written in the form 8q + 1 for some integer q.

Question

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

Solution 

Let the given odd positive integer be written in the form
2n + 1, where n is an integer.

Now, square the given number:

(2n + 1)²
= 4n² + 4n + 1
= 4n(n + 1) + 1

Since n and n + 1 are two consecutive integers, one of them must be even.
Therefore, the product n(n + 1) is even.

Let n(n + 1) = 2q, where q is an integer.

Then,

4n(n + 1) + 1
= 4 × 2q + 1
= 8q + 1

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

Conclusion

Therefore, the square of any odd positive integer can always be written in the form 8q + 1 for some integer q.

Hence proved.

Next Question