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1h
comment Example of commutative ring that doesn't satisfy distribution of intersection over addition
@avi It was already hypothesized that $\,\rm z\,$ is cancellable.
2h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael I don't know why you think the issue has something to do with cancellation. Rather, it has to do with handwaving vs. rigorous justification of crucial. fundamental properties closely related to unique factorization.If you try applying your argument on said example in $\,\Bbb Z[\sqrt{-5}]\,$ then this may help to better understand the points I raise above. Then you can see where the argument breaks down, and determine precisely what special properties of $\,\Bbb Z\,$ are necessary for the argument to go through there.
2h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael Consider applying your argument to $\, w\bar w/2 = 3,\,\ w = 1\!+\!\sqrt{-5}\,$ in $\,\Bbb Z[\sqrt{-5}],\,$ a non-UFD. Does your argument incorrectly conclude that $\, w\mid w\bar w/2,\,$ i.e. $\,2\mid \bar w?\ $
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
Remark that prime $\, n\nmid k,\ n\mid k\,{n\choose k}\,\Rightarrow\, n\mid {n\choose k}\,$ is a consequence of Euclid's Lemma, or unique factorization / FTA (this should be explicitly mentioned since it is essential to the proof).
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael There are many ways it can be proved because there are many properties equivalent to Euclid's Lemma / uniqueness of factorization. But I still have no clue how you intend to complete the above sketch into rigorous proof.
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
But I should not have to guess what you intended. A mathematical proof should not leave such serious doubts.
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael If I had to guess, I would say that perhaps the following is what was intended $\tag*{}$ Lemma $\ \ nj/k \in \Bbb Z,\ (n,k)=1\,\Rightarrow\, n\mid nj/k$ $\tag*{}$ Proof $\ $ By Euclid $\,(n,k)=1,\ k\mid nj\,\Rightarrow\, k\mid j,\ $ so $\,j/k = i\in\Bbb Z,\,$ so $\,n\mid ni=nj/k\ $ $\tag*{}$
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael There is so much handwaving in the answer that I do not know where to begin to debug it. You apparently intend to use some properties of fractions (that depend on unique factorization), but it is not clear precisely what properties you are implicitly using (there are many).
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael Now you have confirmed that the "proof" is not incomplete but, rather, incorrect. So I will downvote this answer. If you fix it then I will be happy to remove the downvote.
3h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@Michael You are using "intuitive" consequences of unique factorization without rigorously justifying them. One of the major goals of a course in elementary number theory to to make rigorous such prior intuition. When one reads a proof like the above it is impossible to know if the (intended) proof is ether incorrect or incomplete. Proofs should not leave such doubts.
4h
comment Why is $ {n\choose k} \equiv 0 \pmod n$ if $n$ is prime?
@NumThcurious While the above idea is correct, much more needs to be said to obtain a rigorous proof. The proof requires unique factorization (or some equivalent such a Euclid's Lemma), so a proper proof must explicitly mention these properties. Better to use LeGrandDODOM's proof, and explicitly write: prime $\, n\nmid k,\ n\mid k{n\choose k}\,\Rightarrow\, n\mid {n\choose k}\ $ by Euclid or unique factorization.
4h
revised Example of commutative ring that doesn't satisfy distribution of intersection over addition
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4h
answered Example of commutative ring that doesn't satisfy distribution of intersection over addition
5h
answered Show that the equation $x^2\equiv a \pmod n$ is solvable $\iff$ $a^{\phi (n)\over 2}\equiv 1\pmod n$.
7h
comment Show that the equation $x^2\equiv a \pmod n$ is solvable $\iff$ $a^{\phi (n)\over 2}\equiv 1\pmod n$.
You mean $\,\phi(n)\,$ divides $\,t\phi(n)/2,\,$ not reversely (as written).
1d
comment Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$
@Nagase Yes. It was not clear to me precisely what your context was. Are you working in some formal system?
1d
awarded  Nice Answer
1d
revised Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$
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1d
revised Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$
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1d
answered Proving that $a \dot{-} (b+1) = (a \dot{-} b) \dot{-} 1$