Prove the following statement.

The square of any odd integer has the form 8m+1 for some integer m.

Step-by-step explanation:

As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma, a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

Case 1 : - When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1) ²

= 16q² + 8q + 1

= 8 (2q² + q) + 1

= 8m + 1, where m = 2q² + q

Case 2 : - when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3) ²

= 16q² + 24q + 9

= 8 (2q² + 3q + 1) + 1

= 8m + 1, where m = 2q² + 3q + 1

Now,

We can see that at every odd values of r, square of a is in the form of 8m + 1.

Also we know, a = 4q + 1 and 4q + 3 are not divisible by 2 means these all numbers are odd numbers.

Hence, it is clear that square of an odd positive is in form of 8m + 1