Find two consecutive odd positive integers, sum of whose squares is 290

Answer

To find two consecutive odd positive integers whose squares sum to 290, we can follow these steps:

### Step 1: Define the integers
Let the first odd integer be \( x \). Since we are looking for consecutive odd integers, the next odd integer will be \( x + 2 \).

### Step 2: Set up the equation
According to the problem, the sum of the squares of these integers is 290. Therefore, we can write the equation:
\[ x^2 + (x + 2)^2 = 290 \]

### Step 3: Expand the equation
Now, we need to expand the equation:
\[ x^2 + (x^2 + 4x + 4) = 290 \]
This simplifies to:
\[ 2x^2 + 4x + 4 = 290 \]

### Step 4: Rearrange the equation
Next, we rearrange the equation to set it to zero:
\[ 2x^2 + 4x + 4 - 290 = 0 \]
This simplifies to:
\[ 2x^2 + 4x - 286 = 0 \]

### Step 5: Simplify the equation
We can divide the entire equation by 2 to make it simpler:
\[ x^2 + 2x - 143 = 0 \]

### Step 6: Factor the quadratic equation
Now, we need to factor the quadratic equation \( x^2 + 2x - 143 = 0 \). We look for two numbers that multiply to \(-143\) and add to \(2\). The numbers are \(13\) and \(-11\). Therefore, we can factor the equation as:
\[ (x + 13)(x - 11) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x + 13 = 0 \) → \( x = -13 \) (not a positive integer)
2. \( x - 11 = 0 \) → \( x = 11 \) (this is a positive integer)

### Step 8: Find the consecutive odd integer
The first integer is \( x = 11 \) and the second consecutive odd integer is:
\[ x + 2 = 11 + 2 = 13 \]

### Conclusion
The two consecutive odd positive integers are \( 11 \) and \( 13 \).

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