bio  website  math.haifa.ac.il/~seva 

location  Israel  
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20h

comment 
A combinatorial and number theoretical problem
It is not clear exactly what are you seeking: the number of ways will definitely depend on the specific choice of your numbers. You should restate your question and ask it at an appropriate site, like MathStackExchange. 
Nov 11 
answered  Existence of arithmetic function satisfying a certain property 
Nov 11 
revised 
Is ${\rm lcm}\{{\rm ord}_p(q)\colon q\mid p1,\ q>2\}>\sqrt p\ \,$?
added 70 characters in body; edited title 
Nov 10 
awarded  Nice Question 
Nov 10 
asked  Is ${\rm lcm}\{{\rm ord}_p(q)\colon q\mid p1,\ q>2\}>\sqrt p\ \,$? 
Nov 9 
awarded  Enlightened 
Nov 9 
awarded  Nice Answer 
Oct 29 
comment 
Why $\gcd \{ {\rm ord}_p(q)\colon q\mid p1 \}$ likes to be large?
Thanks, Lucia, your remark is quite insightful (though it seems difficult to prove anything along these lines). 
Oct 29 
revised 
Why $\gcd \{ {\rm ord}_p(q)\colon q\mid p1 \}$ likes to be large?
(Some) typos fixed 
Oct 29 
asked  Why $\gcd \{ {\rm ord}_p(q)\colon q\mid p1 \}$ likes to be large? 
Oct 13 
comment 
Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?
@BenjaminDickman: Thanks for reminding me about mathoverflow.net/a/165441/22971; I will check how relevant it is. 
Oct 10 
awarded  Nice Question 
Oct 10 
awarded  Popular Question 
Oct 10 
comment 
Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?
@The Masked Avenger: I've just made the computation  nothing to write home about. (I have troubles fitting the results into this comment, but they are absolutely nonilluminating, anyway.) 
Oct 10 
comment 
Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?
@Felipe: having thought for a little while, I don't see any immediate relation. It has to do with $y^{2(n1)}x^{2(n1)}$ being quadratic residues though. 
Oct 10 
asked  Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold? 
Oct 10 
awarded  Yearling 
Sep 30 
awarded  Explainer 
Sep 26 
awarded  Popular Question 
Sep 26 
comment 
List of integers without any arithmetic progression of n terms
@Timothy: my wild guess is that considering progressions with just two carefully chosen differences ($n$ and $n1$?) may suffice. I think the first step would be to verify this numerically. 