### Theory:

Numbers between square numbers:
Numbers between $$1^2$$ and $$2^2$$ are $$2$$, $$3$$.

So, there are $$2 = 2(1)$$ numbers between $$1^2$$ and $$2^2$$.

Numbers between $$2^2$$ and $$3^2$$ are $$5$$, $$6$$, $$7$$, $$8$$.

So, there are $$4 = 2(2)$$ numbers between $$2^2$$ and $$3^2$$.

Numbers between $$3^2$$ and $$4^2$$ are $$10$$, $$11$$, $$12$$, $$13$$, $$14$$, $$15$$.

So, there are $$6 = 2(3)$$ numbers between $$3^2$$ and $$4^2$$.
In general, we can say that there are $$2n$$ non-perfect square numbers between the squares of the numbers $$n$$ and $$(n + 1)$$.
Summing up odd natural numbers:
Odd numbers are $$1$$, $$3$$, $$5$$, $$7$$, $$9$$, $$11$$, $$13$$, $$15$$, ...

First odd number $$=$$ $$1 = 1^2$$

Sum of first two odd numbers $$=$$ $$1 + 3 = 4 = 2^2$$

Sum of first three odd numbers $$=$$ $$1 + 3 + 5 = 9 = 3^2$$

Sum of first four odd numbers $$=$$ $$1 + 3 + 5 + 7 = 16 = 4^2$$

Sum of first five odd numbers $$=$$ $$1 + 3 + 5 + 7 + 9 = 25 = 5^2$$

….

Sum of first $$n$$ odd numbers $$=$$ $$1 + 3 + 5 + 7 + 9 + 11 + … = n^2$$
The sum of the first $$n$$ consecutive odd natural numbers is $$n^2$$.
Summing up two consecutive natural numbers results in an odd square:
Let us take any odd natural number, say $$9$$.

The square of $$9$$ is $$81$$.

To find the sum of two consecutive natural numbers equal to the odd square, we can use the below formula.

${a}^{2}\phantom{\rule{0.147em}{0ex}}=\frac{{a}^{2}-1}{2}+\phantom{\rule{0.147em}{0ex}}\frac{{a}^{2}+1}{2}$

Now, let's take $$a = 9$$.

Substitute the value of $$a$$ in the above formula.

${9}^{2}\phantom{\rule{0.147em}{0ex}}=\frac{{9}^{2}-1}{2}+\phantom{\rule{0.147em}{0ex}}\frac{{9}^{2}+1}{2}$

$81=\frac{80}{2}+\frac{82}{2}$

$$81 = 40 + 41$$

The numbers $$40$$ and $$41$$ are consecutive.

Therefore, $$81$$ can be written as the sum of $$40$$ and $$41$$.
The square of any odd number can be written as the sum of two consecutive natural numbers.
Product of two consecutive even or odd natural numbers:
Let us take two consecutive even natural numbers, $$4$$ and $$6$$.

$$4 \times 6 = 24 = 25 - 1 = 5^2 - 1$$

Here, $$4$$ can be written as $$5 - 1$$, and $$6$$ can be written as $$5 + 1$$.

So, $$(5 - 1) \times (5 + 1) = 5^2 - 1$$.

Now, let us take two consecutive odd natural numbers, $$7$$ and $$9$$.

$$7 \times 9 = 63 = 64 - 1 = 8^2 - 1$$

Here, $$7$$ can be written as $$8 - 1$$, and $$9$$ can be written as $$8 + 1$$.

So, $$(8 - 1) \times (8 + 1) = 8^2 - 1$$.
In general, we can say that $$(a - 1) \times (a + 1) = a^2 - 1$$.