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$$\exists n\forall m ((n+8) \mid ((m+1)^{n+8} - m -1))$$

Where $n$ and $m$ are natural numbers.

How can I determine whether this expression is true or false?

What I have done thus far is the following ...

I rewrite it as $\exists n\forall m (\frac{(m+1)^{n+8} - m -1)}{n+8} \in \mathbb{N})$

Then I can write a series using $m = 0,1,2,3 ...$ which gives me

$$0, \frac{2^{n+8} - 2}{n+8}, \frac{3^{n+8} - 3}{n+8}, \frac{4^{n+8} - 4}{n+8}, \frac{5^{n+8} - 5}{n+8}, ...$$

At this point I am stuck, I have to figure out whether there exists an $n$ such that all of these fractions become integers, I don't see how I can find such an $n$

Arnold Doveman
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2 Answers2

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The statement is true : If $n+8$ is a prime number, we have $$(m+1)^{n+8}\equiv m+1\mod (n+8)$$ for all $m$ due to Fermat's little theorem.

Peter
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  • But Fermat's little theorem also requires that $m+1$ is not divisible by $n+8$. But for any given value of $n$, I can find a $m$ such that $n+8$ divides $m+1$. Forexample if $n=3$, and I choose $m = 21$, I get from Fermat's little theorem that $22^{10} \equiv 1 (\bmod 11)$, which is false since $22$ is divisible by $11$ – Arnold Doveman Oct 12 '17 at 09:18
  • @ArnoldDoveman The version $x^p\equiv x\mod p$ is true for all $x$ including the $x$ with $p|x$. The version $x^{p-1}\equiv 1\mod p$ is only true when $p$ does not divide $x$. In the answer, I used the first version covering the case that $n+8$ divides $m+1$. Anyway, the case that $n+8$ divides $m+1$ is trivial because we just have $0\equiv 0\mod (n+8)$ – Peter Oct 12 '17 at 10:02
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It's true.

Take $n=3$ and use the Fermat's little theorem.

Let $m+1=a$.

We need to prove that $$a^{11}-a$$ is divisible by $11$ for all natural $a$.

By the way, you can check it by yourself without Fermat's little theorem.

Michael Rozenberg
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  • Something doesnt add up. If $n=3$, the statement will hold true if $a^{10} \equiv 1 (\bmod 11)$. This is Fermat's little theorem. But this theorem also requires that $a = m+1$ is not divisible by 11. But if I choose $m = 21$ so that $a=22$. Then I get $22^{10} \equiv 1 (\bmod 11)$ which doesn't hold, which also means $22^{11} \equiv 22 (\bmod 11)$ doesnt hold true – Arnold Doveman Oct 12 '17 at 09:13
  • Arnold Your problem is "There is natural $n$ for which for all natural $m$ we have $(m+1)^{n+8}-(m+1)$ is divisible by $n+8$. If $n=3$ and $m+1=a$ then for all natural $m$ we have $a^{11}-a$ is divisible by $11$, which you wished. $22^{11}-22$ is divisible by $11$ because $22$ is divisible by $11$. – Michael Rozenberg Oct 12 '17 at 09:32