Question

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

Hint:

For consecutive odd (or even) numbers, always represent them as \( x \) and \( x + 2 \) to form a solvable quadratic equation.

Answer 1

Step 1: Let the integers be defined.
Let the two consecutive odd positive integers be \( x \) and \( x + 2 \).

Step 2: Form the equation.
According to the question, \[ x^2 + (x + 2)^2 = 290 \]
Step 3: Simplify the equation.
\[ x^2 + x^2 + 4x + 4 = 290 \Rightarrow 2x^2 + 4x + 4 - 290 = 0 \Rightarrow 2x^2 + 4x - 286 = 0 \Rightarrow x^2 + 2x - 143 = 0 \]
Step 4: Solve the quadratic equation.
\[ x^2 + 2x - 143 = 0 \] Using factorization: \[ (x + 13)(x - 11) = 0 \Rightarrow x = -13 \text{ or } x = 11 \] Since we need positive integers, \( x = 11 \). Thus, the two consecutive odd positive integers are \( 11 \) and \( 13 \).
Step 5: Verification.
\[ 11^2 + 13^2 = 121 + 169 = 290 \] Hence, verified.
Step 6: Conclusion.
The required consecutive odd positive integers are \( 11 \) and \( 13 \).