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How can I prove that for all $n\in\mathbf{N}$ that $6 | n^5 + 5n$?

I tested for $n = 2$ and got $6 | 32 + 10 = 42$.

Sergio Parreiras
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user3704516
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4 Answers4

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$$n^5+5n=6n+(n-1)n(n+1)(n^2+1)$$

Now $6=3!$ divides the product of any three consecutive integers(why?)

lab bhattacharjee
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1) Show that $2|n^5+5n$.

2) Show that $3|n^5 + 5n$.

If we can show both, separately, it has to follow that $6=2*3|n^5 + 5n$.

1)$n^5 + 5n = n(n^4+5)$. If $n$ is even so is $n(n^4+5)$. If $n$ is odd then $n^5$ is odd and $n^5 + 5$ is even and $n(n^4+5)$ is even.

So $n^5 + 5n$ is even.

2) $n = 3m + i$ for some $m$ and $i = 1, 0, $ or $-1$.

a) $i = 0; n = 3m$ then $3|n$ and $3|n(n^4 + 5)$.

b) $i = \pm 1$ then

$n^4 + 5$ = $(3m \pm 1)^4 + 5$

$ = (3^4m^4 \pm 4*3^3m^3 + 6*3^2m^2 \pm 4*3m + 1) + 5$

$= 3^4m^4 \pm 4*3^3m^3 + 6*3^2m^2 \pm 4*3m + 6$

$= 3(3^3m^4 \pm 4*3^3m^3 + 6*3m^2 \pm 4*m + 2)$

So $3|n^4 + 5$ and $3|n^5 + 5n$.

====

So both $2|n^5+5n$ and $3|n^5 + 5n$ so $6|n^5+5n$.

fleablood
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  • sigh.. I knew I would make that typo.... sigh.... I just didn't know I would make it so often – fleablood Oct 18 '16 at 17:52
  • @fleablood I got confused with your first sentence (line three). e.g. 2 divides 30. 6 divides 30. but 2 * 6 = 12 doesn't divide 30. isn't it? Can you explain? –  Oct 18 '16 at 19:15
  • 6 is a different number than 3. I said this was true for 2 and 3. I didn't say it was true for any $a$ and $b$. – fleablood Oct 18 '16 at 19:38
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    If $a|w$ and $b|w$ then $\frac{ab}{\gcd(a,b)}|w$. If $\gcd(a,b) = 1$ then $a|w$ and $b|w$ implies $ab|w$. But only if $\gcd(a,b) = 1$. – fleablood Oct 18 '16 at 19:40
  • @fleablood ah, ok thanks - now I see that general theorem –  Oct 18 '16 at 19:47
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Work modulo 6:

$$\begin{align*} \text{if $n \equiv 0 \mod 6$,}&\qquad n^5+5n\equiv 0,\\ \text{if $n \equiv 1 \mod 6$,}&\qquad n^5+5n\equiv 1+5=6\equiv0,\\ \text{if $n \equiv 2 \mod 6$,}&\qquad n^5+5n\equiv 32+10\equiv2+4\equiv0,\\ \text{if $n \equiv 3 \mod 6$,}&\qquad n^5+5n\equiv 3+3\equiv0,\\ \text{if $n \equiv 4 \mod 6$,}&\qquad n^5+5n\equiv 4+2\equiv0,\\ \text{if $n \equiv 5 \mod 6$,}&\qquad n^5+5n\equiv 5+1\equiv0.\\ \end{align*} $$

So $n^5+5n$ is always a multiple of 6.

zar
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Use modular arithmetic and solve for the finite number of cases $n \equiv 0, 1, 2, 3, 4, 5 \pmod{6}$.

For example, $4 \cdot 4 = 16 \equiv 4 \pmod{6}$, so that $4^5 \equiv 4 \pmod{6}$, and $n^5 + 5n \equiv 4 + 20 \pmod{6} \equiv 24 \pmod{6} \equiv 0 \pmod{6}$.

The five other cases are similar.

  • By the way, when I said "similar," I really meant it. In each case, $n^5 \equiv n \pmod{6}$, so that $n^5 + 5n \equiv 6n \pmod{6} \equiv 0 \pmod{6}$.

    This is also true for $n^3 + 5n$, $n^5 + 5n$, $n^7 + 5n$, $n^{11} + 5n$ and so on. The reason is Fermat's Little Theorem.

    Note that it does not work, however, for $n^2 + 5n$. Fermat's little theorem only applies to $n^p + 5n$ when $(p, n) = 1$, i.e., when $p$ and $n$ are coprime. Thus, for $p = 2, 3, 5$, we have to check the single respective case $n = 2, 3, 5$ by hand. However, for all other primes, it holds.

     –  Oct 20 '16 at 14:25