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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. Else $k$ is even, 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?

Bill Dubuque
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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
  • $,n+1\mid f(n)\iff n+1\mid f(-1) = (-1)^k+1$ by divisibility mod reduction in the linked dupe. $\ \ $ – Bill Dubuque Sep 30 '24 at 22:57

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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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Set base cases $k = 0,1$ so $n + 1 \mid n^{k} + 1$ for some odd $k$ and $n+1 \nmid n^k +1$ for some even $k$.

$n^{k+2} - n^{k} = n^{k}(n^2 - 1) = n^{k}(n-1)(n+1)$ and

$n + 1 \mid (n^{k+2} - n^k)+(n^k +1) = n^{k+2}+1$ for odd $k$ and $n+1 \nmid n^{k+2}+1$ for even $k$.

Derek Luna
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  • This handles only one direction, i.e. it is true for all odd $k$. You still need to prove that it is false for even $k$. Your inductive step shows $,P(k+2)\iff P(k)\ $ (where $P(k):= n+1\mid n^k+1),,$ which enables us to lift its truth at $k=1$ up to all odds, and its non-truth at $k=0$ up to all evens, i.e. we need two base cases $,k=0,1,$ to handle both parity cases. – Bill Dubuque Oct 01 '24 at 01:53
  • Essentially the induction shows $P(k)\iff P(k\bmod 2).,$ It's much clearer in arithmetical vs. divisibility language: $\bmod x+1!:\ x\equiv -1\Rightarrow x^2\equiv 1 \Rightarrow x^k\equiv x^{k\bmod 2},$ by mod order reduction $\ \ $ – Bill Dubuque Oct 01 '24 at 01:53
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Oct 01 '24 at 01:54
  • @BillDubuque I don't believe everything is a duplicate of its generalization, especially in regards to the purposes of the problem (whose purpose is often not for the sake of generalization) . This answer is different from the others. Furthermore, I find the extent of your generalizations are often an overreaction to your distaste for ad-hoc methods. And clarity is subjective; I don't find the language in your second comment (or in many of your answers) more clear than mine. I wouldn't have answered if I felt it was similar to the others. – Derek Luna Oct 01 '24 at 09:24
  • I believe that policy is for actual dupes, not answers/questions that aren't to the level of generality you desire (whose extended generality can be coincidentally related to some answer in a completely different question). The Search for Duplicates seems to be for minor variations, not attaching every question to its generalization (especially of the personal variety). – Derek Luna Oct 01 '24 at 09:29
  • Re: your edit: there is no need to repeat the proof for even $k$. As I explained above we can prove it is true for $k =0,1$ (base cases) then use as inductive step that it is true at $k$ iff it is true at $k-2$, since $,n+1\mid n^k+1 \iff n\mid n^{k-2}+1,,$ by $n+1$ divides their difference $ = n^k-n^{k-2} = n^{k-2}(n^2-1)$. Essentially this inductive proof shows $!\bmod n^2-1!:\ n^k+1\equiv n^r+1,\ r = k\bmod 2 = 0,1,$ (base cases), i.e. $,n^k+1 = n^r+1 + q(n^2-1),,$ some $,q\in\Bbb Z,,$ so $,n+1\mid n^k+1\iff n+1\mid n^r+1,,$ by $,n+1\mid n^2-1.,$ (= mod order reduction above) – Bill Dubuque Oct 01 '24 at 20:11
  • I think I understand this better. The base cases extend the reasoning everywhere and the separation by parity isn't necessary. – Derek Luna Oct 01 '24 at 21:19
  • It's a form of induction by modular (remainder) base cases, i.e. if $,P(k)!\iff! P(k!-!m),$ is true for all $,k\ge m,$ then the truth of $P$ is the same at $,k,,k!-!m,,k!-!2m,\ldots, k!-!jm = k\bmod m = $ remainder left by $,k\div m,,$ so $P$ is true for all naturals $!\iff! P$ is true for all $!\bmod m,$ remainders $,k=0,1,\ldots,m!-!1$ (base cases). OP is the (parity) case $,m=2,,$ and $,P(k):=n+1\mid n^k+1.\ \ $ – Bill Dubuque Oct 01 '24 at 22:05