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For that question, I can use modular arithmetic to prove divisibility. Look at the following: $$n \equiv-1\mod(n+1)$$ raising to $k^{th}$ power, if $k$ is odd, then $$n^k \equiv(-1)^k \equiv-1\mod(n+1)$$ hence $$n^k+1 \equiv0\mod(n+1)$$ as desired. On the other hand, if $k$ is odd, then $$n^k \equiv(-1)^k \equiv1\mod(n+1)$$ hence $$n^k+1 \equiv2\mod(n+1)$$

However, is it possible to come up with a divisibility proof? i.e. if $(n+1)k = (n^k+1)$, what is $k$?? I applied a long division but found $(n^{k-1}-n^{k-2}+n^{k-3}-n^{k-4}...)$ which I am suspicious about as I found in my instructor notes that if $k$ is odd, then $(n^{k}+1)=(n+1)(n^{k-1}-n^{k-2}...-n+1)$. This solution, the instructor's one, feels reasonable as when multiplying these terms, the result is $n^k+1$. What about mine?, how to connect it to the parity of $k$ as well?

Maged Saeed
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    You might want to remember that modular arithmetic is useful in large measure because it encodes information about divisibility, so a proof using modular arithmetic is, in this sense, a divisibility proof. – Mark Bennet Oct 11 '18 at 14:07
  • https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd – lab bhattacharjee Oct 11 '18 at 14:12
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    Why do you seek to eliminate congruences? One of the great advantages of congruence proofs is that they may greatly simplify divisibility proofs,. Eliminating congruences usually serves to greatly obfusctae the arithmetical essence of the matter (above that $-1$ has order $2$ so its powers have simple form). – Bill Dubuque Oct 11 '18 at 15:11
  • I want to know the result when dividing these number. Please re-read the question. I have stated that after proving the divisibility by congruences. – Maged Saeed Oct 11 '18 at 15:15

3 Answers3

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Write $n^k+1=(n+1)q(n)+r(n)$, where $\deg(r)<\deg(q)$.

If $n+1\mid n^k+1$ then $r(n)=0$, so $n^k+1=(n+1)q(n)$. Letting $n=-1$ we get $(-1)^k+1=0$, so $k$ is odd. If $k$ is odd then $n=-1$ is a root, so $n+1\mid n^k+1$.

cansomeonehelpmeout
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$((n+1)-1)^k +1=$

$(n+1)^k + k(-1)(n+1)^{k-1}+......$

$..+k(n+1)(-1)^{k-1}+(-1)^k +1.$

All terms, except the last term $(-1)^k$, in the binomial expansion have a factor $(n+1)$.

For odd $k$: $(-1)^k +1=0.$

Peter Szilas
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For your division, what is the last term? The general term will be $(-1)^{r-1}n^{k-r}$ and you will end when $r=k$ with $(-1)^{k-1}n^0=(-1)^{k-1}$.

Here you get $+1$ when $k$ is odd and $-1$ when $k$ is even. All that remains is to compute the remainder.

One way of doing this is simply to use the remainder theorem - if $$p(n)=(n+1)q(n)+r$$ using polynomial division, then $p(-1)=r$, and apply it to the polynomial $p(n)=n^k+1$

Indeed, it is not necessary to compute $q(n)$ specifically to use this result.

Mark Bennet
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