First, resolve the given number into prime factors. Group the identical factors in pairs and then take one from them to find the square root.

Now, 324 = 2 x 2 x 3 x 3 x 3 x 3
= 2² x 3² x 3 
= (2 x 3 x 3)²
\(\therefore \sqrt { 324 } =\sqrt { \left( 2\times 2\times 3 \right) ^{ 2 } } \) 
= 2 x 3 x 3
= 18

We proceed as follows:
Step 1: 64 – 1 = 63
Step 2: 63 – 3 = 60
Step 3: 60 – 5 = 55
Step 4: 55 – 7 = 48
Step 5: 48 – 9 = 39
Step 6: 39 – 11 = 28
Step 7: 28 – 13 = 15
Step 8: 15 – 15 = 0
We have subtracted successive odd numbers (starting from 1) repeatedly from the given number 64. We get zero in the 8^th step.
Therefore \(\sqrt { 64 } =8\). If we don't get zero, then the given number is not a perfect square.
Given that 685584 is a perfect square number. Is it possible for us to fi nd out its square
root by the method of successive subtraction of odd numbers?
Yes, we can, but it will be time consuming and tedious. So we need to consider other methods of fi nding the root of square numbers.

We have, 8² = 8 × 8 = 64, 15² = 15 × 15 = 225 and 17² = 17 × 17 = 289
We find, 8² + 15² = 64 + 225 = 289 = 17²
Thus, 8² + 15² = 172
∴ (8, 15, 17) is a Pythagorean triplet

Create a 3 × 3 square below and to the left of 345 (red). Make a diagonal in each of the 9 squares created. Place the product value, say, 5 × 3 = 15, 1 in the upper part and 5 in the lower part of the square. After filling all the squares, add the numbers through coloured diagonals. The arrow indicated numbers from the left 119025 square of 345.
ஃ345² = 119025.

Solution:
Here, there are 50 odd numbers from 1 to 99.
The sum of the fi rst n consecutive odd natural numbers = n²
The sum of the fi rst 50 consecutive odd natural numbers = 50²
= 50 × 50 = 2500