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What is the largest integer which must evenly divide all integers of the form $n^5-n$ ?

I am stuck on this problem,I don't know how to approach this.

Some scribble I've tried is:

Given that $n^5-n \equiv 0 \mod x $,I have $\left(n- \cfrac{1}{2} \right)^2 \equiv \cfrac{1}{4} \mod x \implies n \equiv 1 \mod x \implies x\cdot q =n-1 $ for some integer $q$.

I know that $n^5-n$ is always a multiple of $5$ by Fermat's little theorem so $x$ is some multiple of $5$ but after that I don't know what to do.

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    $\begingroup$ It's a good start. How about $n^5 -n $ modulo some other (small) integers? $\endgroup$ – hardmath Jan 22 '16 at 18:12
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    $\begingroup$ $2^5-2$ dramatically limits what other numbers you need to check. $\endgroup$ – Arthur Jan 22 '16 at 18:15
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    $\begingroup$ "I know that $n^5−n$ is always a multiple of $n^5 -n$ by Fermat's little theorem so x is some multiple of 5 but after that I don't know what to do." Well, that's the most important part. next thing to not is that $n^5 - n= n(n^4 - 1)= n(n^2 - 1)(n^2 + 1) = n(n+1)(n-1)(n^2 + 1)$ so n, n+1, n-1 are three consecutive integers so is a mulitple of 3 and or 2 so 2*3*5 = 30 will divide all. As $2^5 - 2 = 30$ 30 is the largest number that divides all. ... interesting. This means for any 3 consect integers, none div by 5, then the middle one squared + 1 is always div by 5... neat... $\endgroup$ – fleablood Jan 22 '16 at 18:29
  • $\begingroup$ See also math.stackexchange.com/questions/164524/…. $\endgroup$ – lhf Jan 22 '16 at 19:22
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A good start is to look at $2^5-2$ because $x$ must divide it also :

$2^5-2=30=2 \cdot 3 \cdot 5$

You checked that $5 \mid n^5-n$ for every $n$ .

It's also clear that $2 \mid n^5-n$ because they have the same parity .

Finally :

$$n^5-n=(n^3-n)(n^2+1)$$ and from Fermat's theorem $3 \mid n^3-n$ so it also follows that $3 \mid n^5-n$ .

Putting everything together it follows that $x=30$ is the required number .

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  • $\begingroup$ Thanks for the answer. One last doubt :how do I know that I won't get an extra factor from $n^2+1 $ ? $\endgroup$ – Mr. Y Jan 22 '16 at 18:47
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    $\begingroup$ @Mr.Y So you know that $2,3$ and $5$ divides all the numbers and this means that $30$ divides all the numbers .But the number you're searching needs to divide also $$2^5-2=30$$ so these are the only possible factors . $\endgroup$ – user252450 Jan 22 '16 at 18:49
  • $\begingroup$ @ComplexPhy Thanks again,that cleared the confusion ! $\endgroup$ – Mr. Y Jan 22 '16 at 18:51
  • $\begingroup$ Well you won't get another factor of 3 from $n^2 + 1$ but that's coincidental. and if 5 doesn't divide $n^3 - n$ then you will get a factor of 5 out of $n^2 + 1$. And you could of course get all sorts of factors out of $n^2+1$ (if n is odd; you'll get extra factors of 2, etc.). But because $2^5 - 2$ = 30 you know thirty is a max possible required value. Those potential extra factors don't matter. the min possible. required value.... $\endgroup$ – fleablood Jan 22 '16 at 19:12
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Hint: $$n^5-n=n(n-1)(n+1)(n^2+1)$$ The first thing to notice would be that there are three consecutive integers - $(n-1)n(n+1)$. Also checking modulo 5 might help.

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$2^5-2=30=2\cdot3\cdot5=6\cdot5$

$N=n^5-n=(n-1)(n)(n+1)(n^2+1)\Rightarrow N\equiv0 \pmod6$.

$n\equiv 1\pmod5\Rightarrow n-1\equiv 0\pmod5$

$n\equiv 2\pmod5\Rightarrow n^2+1\equiv 0\pmod5$

$n\equiv3\pmod5\Rightarrow n^2+1\equiv 0\pmod5$

$n\equiv4\pmod5\Rightarrow n+1\equiv 0\pmod5$

Consequently the asked number is $\color{red}{30}$

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  • $\begingroup$ Thanks sir for the answer,it provided me other insights about the problem :) $\endgroup$ – Mr. Y Jan 22 '16 at 19:01
  • $\begingroup$ It is a pleasure to help you, Mr Y $\endgroup$ – Piquito Jan 22 '16 at 19:17

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