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It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.

Basically, the issue is how to show that :
$(n-1)(n)(n+1)(n^2+1)$ for $n$ being odd or even, is a multiple of $30$.

Let us take two cases:
(a)$n$ is odd, i.e. of the form $2k+1$, for a suitable integer $k$.
So, $(2k)(2k+1)(2k+2)(2k+1)$, as $n^2+1$ will also be an odd number.
This leads to a final expression of the form: $2k+2$.
But, not able to prove its being a multiple of $30$.

(b) Need not take even case, as the final even form is proved by one term being even; & still not able to prove it being a multiple of $30$.

So, try by smaller values, and that would not serve the cause until use induction (strong case might be needed, as need consider all lower value for the case of $k=1$).

Am I correct, & should proceed by strong induction?

Jaideep Khare
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jitender
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  • Apply the Fermat Theorem for $p=2,3,5$ – Exodd Feb 06 '18 at 14:51
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    The statement is true for $n=0,1,2,3$ and by considering the forward difference operator $\delta: p(x)\mapsto p(x+1)-p(x)$ applied to the polynomial $p(x)=x^5-x$ we have $$ \delta^3 p(x) = \color{red}{30}\cdot(2x^2+6x+5) $$ hence the claim holds by induction on $n$. – Jack D'Aurizio Feb 06 '18 at 14:54
  • @Exodd Just glanced it few times, so please elaborate it. Else, will take me time to do it. – jitender Feb 06 '18 at 14:55

5 Answers5

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Because $(n-1)n(n+1)$ is divided by $6$ and for $n=5k\pm2$ we see that $n^2+1$ is divided by $5$.

For $n=5k$ and $n=5k\pm1$ we see that $(n-1)n(n+1)$ is divided by $5$.

Also, it follows immediately from the Fermat's little theorem for $p=5$.

Michael Rozenberg
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Modulo $5$ we have $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)\equiv n(n^2-1)(n^2-4) =n(n-1)(n+1)(n-2)(n+2)$$ and the product of five successive integers is divisible by $5$, while $n(n^2-1)=n(n-1)(n+1)$ is divisible by $6$.

Mark Bennet
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The way I would do this is to prove that $n^5 -n$ is divisible by $2$, $3$, and $5$, so then it must be divisible by $30 = 2 \cdot 3 \cdot 5$ (because $2,3,5$ are all coprime). $2$ and $3$ are relatively straightforward, since there are three consecutive integers $(n-1)n(n+1)$ in the factorization, and we don't even need to consider the $(n^2+1)$ factor; $5$ is a bit trickier, but if you consider all cases $n \equiv 0,1,2,3,4$ mod 5, you find that one of the four factors $(n-1)n(n+1)(n^2+1)$ is $\equiv 0$ mod 5.

Strong induction is also a valid strategy. It's not so obvious exactly how you'd set it up, but it's doable; e.g., check small cases and then show that if $30 \mid (n-5)^5 - (n-5)$, then $30 \mid n^5 - n$, or something like that.

BallBoy
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  • I hope you meant $\equiv 0 \pmod {30}$. If so, then not only for $n=5$, but also for $n=4$, the same ($0$ residue) equivalence class occurs; as $(4-1)(4)(5)=60, (4)(5)(6)=120$. Please vet the above, and also tell why $(n^2+1)$ factor should be simply ignored. – jitender Feb 06 '18 at 15:49
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    @jitender Edited to (hopefully) make clearer. I meant $\equiv 0$ mod 5, since my strategy is to check separately for divisibility by $2$, $3$, and $5$. If we were checking mod 30 we'd have to check $n \equiv 0,1,2,3,4,5,6,7,\dots,29$ mod 30, which would take a while! The $n^2+1$ factor should not be ignored: it's not necessary to use it when checking divisibility by $2$ and $3$, but it will be necessary to use it when checking divisibility by $5$. – BallBoy Feb 06 '18 at 15:52
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As you cane see $n^5 -n=n(n^2-1)(n^2+1)$ now if $n$ is even then $ 2|n^5 -n$ if n is odd then $n^4-1 $ is even so $n^5-n$ is divisible by 2. Now if you divide any positive integer by 3 the remainders are 0, 1 or 2. therefore you can divide $n^5-n$ by 3. Atlast by Fermat's little theorem $5|n^5-n \Rightarrow 30|n^5-n.$

2

You can rearrange: $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$ $$(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1).$$ Both terms are divisible by $2,3,5$, hence by $30.$

farruhota
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  • Very much thanks for this expansion. However, I don't think that in all similar cases, we can expand into product and sum of simpler terms. – jitender Feb 06 '18 at 16:01