**From New Scientist #2580, 2nd December 2006**

This puzzle is concerned with the equation a² + b² + c² = d² + e² + f², where a, b, c, d, e, and f are different integers between 1 and 9.

I found a solution to that equation. I then altered just one of the integers on each side of that solution to create a second solution. I then altered just one of the integers on each side of the second solution to create a third solution. I then altered just one of the numbers on each side of the third solution to create a fourth solution. All four solutions were different.

If the fourth solution produced a larger sum than the first solution, what was the value of (a² + b² + c²) in each of the four solutions in order?

[enigma1420]

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Solution:The values of a² + b² + c² are 62, 77, 122, 98.The equations are:

(1) 1² + 5² + 6² = 2² + 3² + 7² = 62

(2) 4² + 5² + 6² = 2² + 3² + 8² = 77

(3) 4² + 5² + 9² = 7² + 3² + 8² = 122

(4) 4² + 1² + 9² = 5² + 3² + 8² = 98