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Show that $n^5 - n$ is divisible by $30;$ $\forall n\in \mathbb{N}$

I tried to solve this three-way. And all stopped at some point.

I) By induction: testing for $0$, $1$ and $2$ It is clearly true. As a hypothesis, we have $30|n^5-n\Rightarrow n^5-n=30k$. Therefore, the thesis would $30|(n+1)^5-n-1\Rightarrow (n+1)^5-n-1=30j$. By theorem binomial $(n+1)^5-n-1=n^5+5n^4+10n^3+10n^2+5n+1-n-1$ $=30k+5n^4+10n^3+10n^2+5n.$
It was a little messy and not given to proceed.

II)I tried by Fermat's Little Theorem: $$30|n^5-n\Rightarrow 5\cdot3\cdot2|n^5-n $$ Analyzing each case, we have; clearly $5|n^5-n$ (Fermat's Little Theorem). Now cases $3|n^5-n$ and $2|n^5-n$ I could not develop.

III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that $$gcd(n^5-n,30)=30$$ Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD?

Andrés E. Caicedo
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benjamin_ee
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6 Answers6

12

Try factoring:

$$n^5-n=n(n+1)(n-1)(n^2+1)$$

Now, at least one of three consecutive numbers $n-1,n,n+1$ is divisible by $3$, and at least one is divisible by $2$. If none are divisible by $5$, can you show that this implies that $n^2+1$ is divisible by $5$?

Jared
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  • nice! In absence of inspiration, one can always look at all the results of $n^5-n$ mod $30$, for $n=0,1,\dots,29$. – Denis Jul 24 '13 at 17:40
  • Or you can still look at $n = 0,1$ mod $2$, $n = 0,1,2$ mod $3$, $n = 0,1,2,3,4$ mod $5$, which is not that much work even without factoring. – nikoma Jul 24 '13 at 17:49
  • @nikoma nikoma and dkuper, Yet arrived at the contents of mod Unfortunately, but soon, hopefully studying. – benjamin_ee Jul 24 '13 at 18:10
6

You can easily prove it by induction

$$(n+1)^5-(n+1)=(n^5-n)+5n^4+10n^3+10n^2+5n=(n^5-n)+5n\left( n^3+2n^2+2n+1 \right)$$ $$=(n^5-n)+5n(n+1)\left( n^2+n+1 \right)$$ $$=(n^5-n)+5n(n+1)\left( (n-1)^2-3n\right)$$ $$=(n^5-n)+5n(n+1)(n-1)^2-15n^2(n+1)$$

  • is divisible by $30$ by the inductive assumption
  • $n(n+1)(n-1)$ is the product of three consecutive integers, thus divisible by $2$ and $3$.
  • $n(n+1)$ is even.

Added If you use induction probably the best way to approach the problem is by observing that the inductive step reduces to

$$30|5n^4+10n^3+10n^2+5n=5\left( n^4+2n^3+2n^2+n \right)$$ which is equivalent to

$$6|n^4+2n^3+2n^2+n$$ Treat this as a new induction problem.

Note that the degree of the polynomial decreased by $1$, and the inductive step, if not obvious, is also a similar problem but the degree again one less. Repeat until you get an obvious statement...

N. S.
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3

Method $1:$

$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$ $$=n(n^2-1)(n^2-4)+5n(n^2-1)$$ $$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$

Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer

Method $2:$

Using Fermat's Little Theorem, $n^p\equiv n\pmod p$ for prime $p$

$\implies 5$ divides $n^5-n$

$n^5-n=n(n^2-1)(n^2+1)=(n^3-n)(n^2+1)$

As $3$ divides $n^3-n\implies 3$ divides $n^5-n$

$n^5-n=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$

As $2$ divides $n^2-n\implies 2$ divides $n^5-n$

$\implies $lcm $(2,3,5)$ divides $n^5-n$

Community
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lab bhattacharjee
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1

Factor it .

$$n^5-n=n(n+1)(n-1)(n^2+1)$$

1.) Three consecutive integers will be divisible by 2 and 3 .

2.)If $n^2 - 4$ is divisible by 5 then $n^2+1$ is also divisible by 5.

3.)Put $ n^2 - 4$ in place of $n^2+1$


4.) We get $$ n^5-n=(n+1)(n)(n-1)(n-2)(n+2) $$ Which is a product of 5 consecutive integers . And Hence it is divisible by 5 as well .

Simar
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$$\frac{n^5-n}{30} = 4\binom{n}{5} + 8\binom{n}{4} + 5\binom{n}{3} + \binom{n}{2} -n $$

(In fact, every polynomial which takes values in $\mathbf{Z}$ can be expressed as a sum of binomial coefficients with integer coefficients, as above, so this was not magic.)

Sean Eberhard
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Here's another approach, just to feed the imagination for similar problems.

For the factors $2$ and $3$ use @Jared's factorisation.

For the factor $5$ note that the squares modulo $5$ are $0, \pm 1$ and use $n(n^2-1)(n^2+1)$ (or little Fermat, because $5$ is prime).

The point was to draw attention to the squares, since these are sometimes useful and/or significant.

Mark Bennet
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