n^4-n^2=n x (n-1)n(n+1)

deduce that n^4-n^2 is always divisible by 3 , for any positive integer n.

Any help starting off this question would be appreciated,

thanks

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- Sep 15th 2009, 02:41 AMoffahengaway and chipsNumber theory problem
n^4-n^2=n x (n-1)n(n+1)

deduce that n^4-n^2 is always divisible by 3 , for any positive integer n.

Any help starting off this question would be appreciated,

thanks - Sep 15th 2009, 02:49 AMcoquitao
Work modulo 3. If n is already divisible by 3 you are done. Otherwise, n is either 1 or 2 modulo 3. In the former case you have that (n-1) must be divisible by 3. In the latter, you have that it is (n+1) the factor of your expression that is a multiple of 3. Done.

- Sep 15th 2009, 02:57 AMoffahengaway and chips
not quite with you on this one

- Sep 15th 2009, 03:49 AMcoquitao
right?

Then, study all possibles remainders of**n**upon division by**3**. If remainder is 0, then the factor is a multiple of 3 and you are done. If remainder is 1, then the factor is a multiple of 3 and you are also done. Same with remainder 2 and that'll be it. - Sep 15th 2009, 09:27 AMBruno J.
An easy way to say it is that out of three consecutive integers, there is always one divisible by three.

- Sep 16th 2009, 03:07 PMcoquitao
Well, I didn't want to put things that way, esteemed

*Bruno*, 'cause in a sense that's exactly what our friend offahengaway-and-chips has been asked to established. ;) - Sep 16th 2009, 03:20 PMmamen
I think you can use Division algorithm here.

ex. If n is an integer, by D.A., n has the following forms:

"n=2k;n=2k+1" only.

try to substitute n=2k;n=2k+1 to n, and see if it is divisible by 3. If not try other forms.

ex. "n=3k; n= 3k+1 ;n= 3k+2" only - Sep 16th 2009, 03:29 PMPaulRS
Just think of .

In case you don't know what that is :

Consider - the set of all ordered tripes of numbers in where each entry is different from the other 2. -

Note that since - the first entry- may be given n+1 different values, fixed x, can take values, and then fixed the first 2, can take values.

You may partition the set in subsets of 6 elements by considering the equivalence relation: if and only if the sequence is a permutation of .

Then it follows that 6 divides