The sum of the squares of two consecutive even integers is 520. What is the larger integer? - Treasure Valley Movers

The sum of the squares of two consecutive even integers is 520. What is the larger integer?
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Math curiosity is thriving among mobile-first users who value clarity and purposeful learning. Algebraic puzzles like this frame abstract equations as real-world questions, sparking engagement far beyond schoolwork. Social media and educational platforms highlight interactive problem-solving, amplifying interest in integer patterns and even number sequences. With increasing demand for mental discipline and screen-based learning, such puzzles serve as accessible entry points into mathematical thinking—quietly positioning themselves at the center of modern numeracy trends.

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How the Sum of the Squares of Two Consecutive Even Integers Equals 520

Let the two consecutive even integers be expressed as ( x ) and ( x + 2 ), where ( x ) is even.
Their squares are ( x^2 ) and ( (x + 2)^2 ).

Adding the squares:
[ x^2 + (x + 2)^2 = 520 ]

Expanding:
[ x^2 + x^2 + 4x + 4 = 520 ]
[ 2x^2 + 4x + 4 = 520 ]

Subtract 520 from both sides:
[ 2x^2 + 4x - 516 = 0 ]

Divide whole equation by 2:
[ x^2 + 2x - 258 = 0 ]

Now solve the quadratic using the quadratic formula:
[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-258)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 1032}}{2} = \frac{-2 \pm \sqrt{1036}}{2} ]

Check perfect square—note: 1036 = 4 × 259, but 259 isn’t a perfect square; however, testing nearby even integers reveals:
Trying ( x = 14 ):
[ 14^2 + 16^2 = 196 + 256 = 452
]
Too low.