Question

Show that the square of any odd integer is of the form 4q + 1, for some integer q.

Answer 1

By Euclid’s division algorithm, we have a = bm + r, where 0 ≤ r < b … (i)

On putting b = 4 in equation (i), we get

a = 4m + r, where 0 ≤ r < 4, i.e., r = 0, 1, 2, 3

If r = 0 `\implies` a = 4m, 4m is divisible by 2 `\implies` 4m is even.

If r = 1 `\implies` a = 4m + 1, (4m + 1) is not divisible by 2.

If r = 2 `\implies` a = 4m + 2 = 2(2m + 1), 2(2 m + 1) is divisible by 2 `\implies` 2(2m + 1) is even.

If r = 3 `\implies` a = 4m + 3, (4m + 3) is not divisible by 2.

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

Now, a² = (4m + 1)² = 16m² + 1 + 8m  ...[∵ (a + b)² = a² + 2ab + b²]

= 4(4m² + 2m) + 1

= 4q + 1

It is a square which is of the form 4q + 1, where q = (4m² + 2m) is an integer

And a² = (4m + 3)² = 16m² + 9 + 24m  ...[∵ (a + b)² = a² + 2ab + b²]

= 4(4m² + 6m + 2) + 1

= 4q + 1

It is a square which is of the form 4q + 1, where q = (4m² + 6m + 2) is an integer.

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