Question

The number of pairs of consecutive positive even integers such that the sum of their squares is \(290\) is:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

For consecutive even integers, represent them as \(2n\) and \(2n+2\). After forming the equation, check whether the obtained value of \(n\) is a positive integer.

Answer 1

The Correct Option is A

Step 1: Represent the consecutive positive even integers.
Let the two integers be 2n and 2n + 2, where n is a natural number.

Step 2: Formulate the equation from the given condition.
The sum of their squares is (2n)² + (2n + 2)² = 290. Expanding: 4n² + 4n² + 8n + 4 = 290 → 8n² + 8n - 286 = 0. Dividing by 2: 4n² + 4n - 143 = 0.

Step 3: Check for integer solutions using the discriminant.
D = 4² - 4(4)(-143) = 16 + 2288 = 2304, so √D = 48. Then n = (-4 ± 48)/8, giving n = 44/8 = 11/2 or n = -52/8 = -13/2. Neither result is a positive integer.

Step 4: Conclude the number of valid pairs.
Since no natural number n satisfies the equation, there exists no pair of consecutive positive even integers fulfilling the condition.

Step 5: Final conclusion.
The number of such pairs is 0.