Question:
The number of pairs of consecutive positive even integers such that the sum of their squares is \(290\) is:
The number of pairs of consecutive positive even integers such that the sum of their squares is \(290\) is:
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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.
Updated On: Jun 18, 2026
- \(0\)
- \(1\)
- \(2\)
- \(3\)
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The Correct Option is A
Solution and Explanation
Step 1: Let the consecutive positive even integers be \(2n\) and \(2n+2\).
Since the integers are consecutive positive even integers, we can write them as \[ 2n \quad \text{and} \quad 2n+2, \] where \[ n\in \mathbb{N}. \]
Step 2: Form the equation using the given condition.
The sum of their squares is \(290\). Therefore, \[ (2n)^2+(2n+2)^2=290. \] Expanding, \[ 4n^2+4n^2+8n+4=290. \] \[ 8n^2+8n+4=290. \] \[ 8n^2+8n-286=0. \] Dividing by \(2\), \[ 4n^2+4n-143=0. \]
Step 3: Check whether the equation has an integer solution.
Using the discriminant, \[ D=b^2-4ac. \] Thus, \[ D=4^2-4(4)(-143). \] \[ D=16+2288. \] \[ D=2304. \] \[ \sqrt{D}=48. \] Hence, \[ n=\frac{-4\pm48}{8}. \] \[ n=\frac{44}{8}=\frac{11}{2} \] or \[ n=\frac{-52}{8}=-\frac{13}{2}. \] Neither value is a positive integer.
Step 4: Conclude the number of pairs.
Since no positive integer value of \(n\) satisfies the equation, there is no pair of consecutive positive even integers whose squares sum to \(290\).
Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]
Since the integers are consecutive positive even integers, we can write them as \[ 2n \quad \text{and} \quad 2n+2, \] where \[ n\in \mathbb{N}. \]
Step 2: Form the equation using the given condition.
The sum of their squares is \(290\). Therefore, \[ (2n)^2+(2n+2)^2=290. \] Expanding, \[ 4n^2+4n^2+8n+4=290. \] \[ 8n^2+8n+4=290. \] \[ 8n^2+8n-286=0. \] Dividing by \(2\), \[ 4n^2+4n-143=0. \]
Step 3: Check whether the equation has an integer solution.
Using the discriminant, \[ D=b^2-4ac. \] Thus, \[ D=4^2-4(4)(-143). \] \[ D=16+2288. \] \[ D=2304. \] \[ \sqrt{D}=48. \] Hence, \[ n=\frac{-4\pm48}{8}. \] \[ n=\frac{44}{8}=\frac{11}{2} \] or \[ n=\frac{-52}{8}=-\frac{13}{2}. \] Neither value is a positive integer.
Step 4: Conclude the number of pairs.
Since no positive integer value of \(n\) satisfies the equation, there is no pair of consecutive positive even integers whose squares sum to \(290\).
Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]
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