The sum of the squares of three consecutive natural numbers is 77. Find the numbers.

Answer

To solve the problem of finding three consecutive natural numbers whose squares sum up to 77, we can follow these steps:

### Step 1: Define the Variables
Let the three consecutive natural numbers be:
- First number: \( x \)
- Second number: \( x + 1 \)
- Third number: \( x + 2 \)

### Step 2: Set Up the Equation
According to the problem, the sum of the squares of these numbers is 77. Therefore, we can write the equation as:
\[ x^2 + (x + 1)^2 + (x + 2)^2 = 77 \]

### Step 3: Expand the Squares
Now, we will expand the squares in the equation:
\[ x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77 \]
This simplifies to:
\[ x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 77 \]

### Step 4: Combine Like Terms
Combining the like terms gives us:
\[ 3x^2 + 6x + 5 = 77 \]

### Step 5: Rearrange the Equation
Now, we will rearrange the equation to set it to zero:
\[ 3x^2 + 6x + 5 - 77 = 0 \]
This simplifies to:
\[ 3x^2 + 6x - 72 = 0 \]

### Step 6: Factor the Quadratic Equation
Next, we can factor out the common factor of 3:
\[ 3(x^2 + 2x - 24) = 0 \]
Now we need to factor the quadratic \( x^2 + 2x - 24 \).

### Step 7: Find Factors
We look for two numbers that multiply to -24 and add up to 2. The numbers are 6 and -4. Therefore, we can write:
\[ (x + 6)(x - 4) = 0 \]

### Step 8: Solve for \( x \)
Setting each factor equal to zero gives us:
1. \( x + 6 = 0 \) → \( x = -6 \) (not a natural number)
2. \( x - 4 = 0 \) → \( x = 4 \) (valid solution)

### Step 9: Determine the Consecutive Numbers
Now that we have \( x = 4 \), we can find the three consecutive natural numbers:
- First number: \( x = 4 \)
- Second number: \( x + 1 = 5 \)
- Third number: \( x + 2 = 6 \)

Thus, the three consecutive natural numbers are **4, 5, and 6**.