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#math trivia for #September2: #245 can be expressed as a^2+b^2+c^2 for positive integers a, b, c in three ways. What are they?

— Burt Kaliski Jr. (@modulomathy) September 3, 2013

The three ways are

• 1²+10²+12²
• 2²+4²+15²
• 8²+9²+10²

The third way is an interesting one because the integers are consecutive. If we let x denote the middle integer, then the sum of the squares would be

(x-1)²+x²+(x+1)² = x²-2x+1+x²+x²+2x+1 = 3x²+2  .

Any number that is two more than three times a square can thus be expressed as the sum of squares of three consecutive integers. To meet the constraint that the integers must be positive, the number must be at least 1²+2²+3² = 14. (The relationship also holds for the numbers 3 and 5, if non-positive integers are allowed.)

Note:  If we relaxed the constraint that the integers must be positive in the present problem, one more way would be possible:  7²+14². This is a nice consequence of the fact that 245 is the product of a sum of squares, 5, and a square, 49.