Problem Statement:

Find the number of ordered pairs of natural numbers (a, b) such that a^2 + b^2 = 36.

Approach:

To find the number of ordered pairs of natural numbers (a, b) that satisfy the equation a^2 + b^2 = 36, we can systematically analyze the possible values of a and b.

Step 1: Determine the range of values for a and b.
Since a and b are natural numbers, they can take values from 1 to √36 (as the maximum value of a or b can be at most √36, as a^2 + b^2 = 36). Therefore, the range of values for a and b is 1 to 6.

Step 2: Enumerate the possible values of a and b.
We can create a table to list all the possible values of a and b which satisfy the equation a^2 + b^2 = 36.

| a | b |
|---|---|
| 1 | 5 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 1 |

Step 3: Count the number of ordered pairs.
From the table, we can see that there are 5 ordered pairs of natural numbers (a, b) that satisfy the equation a^2 + b^2 = 36.

Answer:

Therefore, the number of ordered pairs of natural numbers (a, b) that satisfy the equation a^2 + b^2 = 36 is 5.