Question 390760
We want to show that


{{{x^2 + (x+1)^2 + (x+2)^2 + 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


{{{0^2 + 1^2 + 2^2 + 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)