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

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) (Show Source): You can put this solution on YOUR website!
We want to show that

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

(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)
RELATED QUESTIONS
Show that the sum of 11 consecutive integers is always divisible by 11. Show that the... (answered by tinbar,richard1234)

How many ways are there to re-arrange numbers 1 to 12 on a circle, so that the sum of any (answered by KMST)

Difference between the squares of two consecutive odd integers is always divisible by:- (answered by stanbon)

The product of two consecutive positive integers is always divisible by an integer... (answered by Greenfinch)

The product of two consecutive positive integers is always divisible by an integer... (answered by Alan3354)

the sum of three consecutive integers is three times the second integer. Set up an... (answered by Alan3354)

Show that the sum of any three consecutive numbers is a multiple of 3. Show that the sum (answered by Alan3354)

Show that the sum of three consecutive integers is always equal to 3 times the middle... (answered by Edwin McCravy)

Three consecutive odd integers are such that the square of the third integer is 15... (answered by vleith,Fombitz)