Question

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

Hint:

To solve problems involving the sum of squares of consecutive integers, represent the integers algebraically and use the properties of quadratic equations to solve for the unknown integer.

Answer 1

Let the two consecutive integers be \( x \) and \( x + 1 \). The sum of their squares is given by: \[ x^2 + (x + 1)^2 = 365. \] Expanding \( (x + 1)^2 \): \[ x^2 + (x^2 + 2x + 1) = 365. \] Simplifying: \[ 2x^2 + 2x + 1 = 365. \] Subtract 365 from both sides: \[ 2x^2 + 2x - 364 = 0. \] Now, divide the equation by 2 to simplify: \[ x^2 + x - 182 = 0. \] This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 1 \), and \( c = -182 \). We can solve it using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Substituting the values of \( a \), \( b \), and \( c \): \[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-182)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm \sqrt{729}}{2}. \] Since \( \sqrt{729} = 27 \), we have: \[ x = \frac{-1 + 27}{2} = \frac{26}{2} = 13. \] Thus, the two consecutive integers are \( 13 \) and \( 14 \).
Conclusion: The two consecutive positive integers are \( 13 \) and \( 14 \).