• Dec 7th 2012, 04:58 AM
vanquishmc
The sum of the squares of 3 consecutive integers cannot have remainder -1 on division by 12.

Can anyone help me prove this by contradiction?
• Dec 7th 2012, 05:08 AM
emakarov
The idea of a proof by contradiction is to assume the negation of what you need to prove and derive a contradiction.

0. Assume that there exists some integer n such that the sum of squares of n, n + 1 and n + 2 has remainder -1 when divided by 12.

1. Write the sum of squares of n, n + 1 and n + 2. Simplify.

2. Write the equation saying that the result of step 1 has remainder -1 when divided by 12. Simplify as much as possible.

3. Consider the cases when n is even and when n is odd and show that the equation from step 2 cannot hold in either case. Show that one side is even and the other is odd, or that one side is divisible by 4 and the other is not, or something like this.

If you need more help, post the results of steps 1 and 2.

Edit: Step 3 of my initial solution was incorrect, but it is now updated.
• Dec 7th 2012, 05:15 AM
RBowman