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
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.
$\displaystyle n^{4}-n^{2} = n^{2}(n^{2}-1) = n^{2}(n-1)(n+1),$ right?
Then, study all possibles remainders of n upon division by 3. If remainder is 0, then the factor $\displaystyle n^{2}$ is a multiple of 3 and you are done. If remainder is 1, then the factor $\displaystyle (n-1)$ is a multiple of 3 and you are also done. Same with remainder 2 and that'll be it.
Just think of $\displaystyle \binom{n+1}{3}$.
In case you don't know what that is :
Consider $\displaystyle A = \left\{ {\left( {x,y,z} \right) \in \left\{ {1,..,n+1} \right\}^3 /x \ne y;x \ne z;y \ne z} \right\}$ - the set of all ordered tripes of numbers in $\displaystyle
{\left\{ {1,..,n+1} \right\}}
$ where each entry is different from the other 2. -
Note that $\displaystyle
\left| A \right| = \left( {n + 1} \right) \cdot n \cdot \left( {n - 1} \right)
$ since $\displaystyle x$ - the first entry- may be given n+1 different values, fixed x, $\displaystyle y$ can take $\displaystyle n$ values, and then fixed the first 2, $\displaystyle z$ can take $\displaystyle n-1$ values.
You may partition the set $\displaystyle A$ in subsets of 6 elements by considering the equivalence relation: $\displaystyle \left( {x_1 ,x_2 ,x_3 } \right) \sim \left( {a,b,c} \right)$ if and only if the sequence $\displaystyle a,b,c$ is a permutation of $\displaystyle x_1 ,x_2 ,x_3 $.
Then it follows that 6 divides $\displaystyle \left| A \right|$