# Thread: Number theory problem

1. ## Number 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

2. 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.

3. not quite with you on this one

4. $\displaystyle n^{4}-n^{2} = n^{2}(n^{2}-1) = n^{2}(n-1)(n+1),$ 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.

5. An easy way to say it is that out of three consecutive integers, there is always one divisible by three.

6. 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.

7. Originally Posted by offahengaway and chips
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
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

8. 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 $\displaystyle \left| A \right|$