Find two consecutive positive integers, sum of whose squares are 365.

Options

A. 13,14
B. 12,13
C. 14,15
D. none of these

Explanation

To find two consecutive positive integers whose squares sum up to 365, we can follow these steps:

### Step 1: Define the integers
Let the first positive integer be \( x \). Then, the second consecutive positive integer will be \( x + 1 \).

### Step 2: Set up the equation
According to the problem, the sum of the squares of these two integers is 365. Therefore, we can write the equation:
\[ x^2 + (x + 1)^2 = 365 \]

### Step 3: Expand the equation
Now, we will expand the equation:
\[ x^2 + (x^2 + 2x + 1) = 365 \]
This simplifies to:
\[ x^2 + x^2 + 2x + 1 = 365 \]
Combining like terms gives:
\[ 2x^2 + 2x + 1 = 365 \]

### Step 4: Rearrange the equation
Next, we will rearrange the equation to set it to zero:
\[ 2x^2 + 2x + 1 - 365 = 0 \]
This simplifies to:
\[ 2x^2 + 2x - 364 = 0 \]

### Step 5: Simplify the equation
We can divide the entire equation by 2 to simplify it:
\[ x^2 + x - 182 = 0 \]

### Step 6: Factor the quadratic equation
Now we need to factor the quadratic equation. We are looking for two numbers that multiply to \(-182\) and add to \(1\). The numbers \(14\) and \(-13\) satisfy these conditions:
\[ (x + 14)(x - 13) = 0 \]

### Step 7: Solve for \( x \)
Now we can set each factor equal to zero:
1. \( x + 14 = 0 \) → \( x = -14 \) (not a positive integer)
2. \( x - 13 = 0 \) → \( x = 13 \)

### Step 8: Find the consecutive integers
Since \( x = 13 \), the two consecutive integers are:
- First integer: \( 13 \)
- Second integer: \( 13 + 1 = 14 \)

### Conclusion
The two consecutive positive integers whose squares sum to 365 are \( 13 \) and \( 14 \).

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