SOLUTION: Show that 1 greater than the sum of the squares of any three consecutive integers is always divisible by 3.

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Question 390760: Show that 1 greater than the sum of the squares of any three consecutive integers is always divisible by 3.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We want to show that

x%5E2+%2B+%28x%2B1%29%5E2+%2B+%28x%2B2%29%5E2+%2B+1 is divisible by 3. Instead of expanding the whole mess out, use modular arithmetic. We know that one of the numbers is equivalent to 0 mod 3, another one is congruent to 1 mod 3, and the third number is equivalent to 2 mod 3 (since they're consecutive). Therefore, the equation is equivalent to

0%5E2+%2B+1%5E2+%2B+2%5E2+%2B+1 (mod 3)

= 6 (mod 3). Since 6 mod 3 and 0 mod 3 are equivalent, we conclude that the sum must be divisible by 3.

(If you're unfamiliar with modular arithmetic, you can go to this Wikipedia article: http://en.wikipedia.org/wiki/Modular_arithmetic)