To find the Pythagorean triplet where the smallest number is 8, we will follow these steps:
### Step 1: Understand the Pythagorean Triplet
A Pythagorean triplet consists of three positive integers \( a \), \( b \), and \( c \) such that:
\[ a^2 + b^2 = c^2 \]
where \( c \) is the largest number (the hypotenuse) and \( a \) and \( b \) are the other two sides of the right triangle.
### Step 2: Identify the Smallest Number
Given that the smallest number is 8, we can denote:
\[ a = 8 \]
### Step 3: Find Possible Values for \( b \) and \( c \)
We need to find integers \( b \) and \( c \) such that:
\[ 8^2 + b^2 = c^2 \]
### Step 4: Calculate \( 8^2 \)
Calculating \( 8^2 \):
\[ 8^2 = 64 \]
### Step 5: Set Up the Equation
Now, we can rewrite the equation:
\[ 64 + b^2 = c^2 \]
This can be rearranged to:
\[ c^2 - b^2 = 64 \]
This can be factored using the difference of squares:
\[ (c - b)(c + b) = 64 \]
### Step 6: Find Factor Pairs of 64
The factor pairs of 64 are:
- (1, 64)
- (2, 32)
- (4, 16)
- (8, 8)
### Step 7: Solve for \( b \) and \( c \)
For each factor pair \( (m, n) \):
1. Set \( c - b = m \)
2. Set \( c + b = n \)
From these two equations, we can solve for \( c \) and \( b \):
- Adding the two equations:
\[ 2c = m + n \implies c = \frac{m + n}{2} \]
- Subtracting the first from the second:
\[ 2b = n - m \implies b = \frac{n - m}{2} \]
### Step 8: Check Each Factor Pair
- **For (1, 64)**:
\[ c = \frac{1 + 64}{2} = 32.5 \quad (not \, an \, integer) \]
- **For (2, 32)**:
\[ c = \frac{2 + 32}{2} = 17 \]
\[ b = \frac{32 - 2}{2} = 15 \]
- Triplet: (8, 15, 17)
- **For (4, 16)**:
\[ c = \frac{4 + 16}{2} = 10 \]
\[ b = \frac{16 - 4}{2} = 6 \]
- Triplet: (8, 6, 10) (but 6 is not the smallest)
- **For (8, 8)**:
\[ c = \frac{8 + 8}{2} = 8 \]
\[ b = \frac{8 - 8}{2} = 0 \quad (not \, valid) \]
### Conclusion
The valid Pythagorean triplet where the smallest number is 8 is:
\[ (8, 15, 17) \]
