Number theory problem

**offahengaway and chips** · September 15th 2009, 01:41 AM

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

**coquitao** · September 15th 2009, 01:49 AM

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.

**offahengaway and chips** · September 15th 2009, 01:57 AM

not quite with you on this one

**coquitao** · September 15th 2009, 02:49 AM

$n^{4}-n^{2} = n^{2}(n^{2}-1) = n^{2}(n-1)(n+1),$ right?

Then, study all possibles remainders of n upon division by 3. If remainder is 0, then the factor $n^{2}$ is a multiple of 3 and you are done. If remainder is 1, then the factor $(n-1)$ is a multiple of 3 and you are also done. Same with remainder 2 and that'll be it.

**Bruno J.** · September 15th 2009, 08:27 AM

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

**coquitao** · September 16th 2009, 02:07 PM

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.

**mamen** · September 16th 2009, 02:20 PM

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

**PaulRS** · September 16th 2009, 02:29 PM

Just think of $\binom{n+1}{3}$ .

In case you don't know what that is :

Consider $A = \left\{ {\left( {x,y,z} \right) \in \left\{ {1,..,n+1} \right\}^3 /x \ne y;x \ne z;y \ne z} \right\}$ - the set of all ordered tripes of numbers in $<br /> {\left\{ {1,..,n+1} \right\}}<br />$ where each entry is different from the other 2. -

Note that $<br /> \left| A \right| = \left( {n + 1} \right) \cdot n \cdot \left( {n - 1} \right)<br />$ since $x$ - the first entry- may be given n+1 different values, fixed x, $y$ can take $n$ values, and then fixed the first 2, $z$ can take $n-1$ values.

You may partition the set $A$ in subsets of 6 elements by considering the equivalence relation: $\left( {x_1 ,x_2 ,x_3 } \right) \sim \left( {a,b,c} \right)$ if and only if the sequence $a,b,c$ is a permutation of $x_1 ,x_2 ,x_3$ .

Then it follows that 6 divides $\left| A \right|$

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