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Suppose the smaller of the two even natural numbers is 2x. Then the other one is 2x + 2

The sum of their squares is (2x)2 + (2x + 2)2 = 4x2 + 4x2 + 8x + 4 = 8x2 + 8x + 4

You want 8x2 + 8x + 4 < 1000

that is 8x2 + 8x - 996 < 0

or 2x2 + 2x - 249 < 0

x < {-2 + sqrt[22 - 4*2*(-249)]}/(2*2)

= {-2 + sqrt[4 + 1992]}/(4)

= {-2 + sqrt[1996]}/(4)

= {-2 + 44.68}/4

= 42.68/4

= 21.34

So x = 20

Then, the sum is 202 + 222 = 400 + 484 = 884

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Q: What is the largest 3 digit number that is the sum of the squares of two consecutive even natural numbers?
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