**Solution:** Let a is any integer. According to question every a is of the form

### 8q + 1.

Let, a = 8q + 1

⟹ a^{2} = (8q +1)^{2}

⟹ a^{2} = 64q^{2} + 16q + 1

⟹ a^{2} = 8(8q^{2} + 2q) + 1

Let, k = 8q^{2} + 2q

So, a^{2} = 8q + 1

*Therefore the square of any odd integer is of the form 8k +1.*