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- Jul 8th 2010, 03:35 AMdwsmith2n^3+3n^2+n\equiv 0 (mod 6)

??? - Jul 8th 2010, 04:45 AMaman_cc
Hint:

This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question) - Jul 8th 2010, 07:30 AMSoroban
Hello, dwsmith!

I have a back-door approach to this one . . .

Quote:

As aman-cc pointed out: .

I recognized this is as part of the formula for the sum of consecutive squares:

. .

Since the left side is an integer, the right side must be an integer.

. . That is, is a multiple of 6.

Therefore: . .forinteger values of*all*

- Jul 8th 2010, 07:14 PMsimplependulum
Consider , use the property of product of few consecutive integers .

- Jul 8th 2010, 11:44 PMmelese

This a solution I saw once:

. One of and is divisible by .

;

. Then at least one of , , and is divisible by .

Since and are relatively prime is divisible by .