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,
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.
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.
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