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Sávio
  • Member for 8 years, 9 months
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Twin Primes- Clement conjecture proof
Just for curiosity, there is a generalization of Clement's theorem that appeared in a 2010 iberoamerican university competition (see ciim.uan.edu.co/ciim-2010-pruebas problem 5). Namely, let $n,d > 1$ be integers with $\gcd(n,d!)=1$. Then $n$ and $n+d$ are both primes if and only if $d!d((n-1)!+1)+n(d!-1) \equiv 0 \pmod {n(n+d)}$.
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Invariant complementary sets modulo $p$
I mean, $2 \le k \le p-1$
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Invariant complementary sets modulo $p$
Indeed, the hypothesis $k \equiv 1 \pmod n$ is useless. You just used $1 \le k \le q-1$, right? Thanks!
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Invariant complementary sets modulo $p$
Ok, I'm sorry. I'm a kind of "math-forums freshman".
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Invariant complementary sets modulo $p$
$p$ instead $q$ in some places
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Geometric progression modulo p
Ok, I'll post a new question. The first question was not really what I needed, it would just easily (but wrong) imply what I need. Well, to discover $s$ is an annoying handwork (I cannot programming), so I confess I didn't check many cases ($p=11,n=5$ only). The second question also implies what I need, but not too easy. I checked some cases by hand ($(n,p,k,s) = (5,11,6,4), (5,31,\{6,11,16,21,26\},16), (7,29,\{8,15,22\},\{7,16\})$) and it really does not seem to be false. An obvious observation is that $2 \le s \le \min\{k-1,q-k\}$.
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Geometric progression modulo p
Adding a new related question
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Number of classes $\pmod p$ represented by $b_1s^{n-1} + \dots + b_n$ where $ord_p(s) = n$
So, the sequences with at least 3 classes $0$, one class $7$ and one class $3$ satisfy what I want. I'd like to show that for every sequence (with $2n-1$ elements such that all classes $\pmod p$ are represented by at most $n-1$ elements) we can take a subsequence $b_1,\dots,b_n$ and valuate it on $c_1s^{n-1} + \dots + c_{n-1}s + c_n$ (where $\{b_1,\dots,b_n\} = \{c_1,\dots,c_n\}$) returning $n+1$ distinct elements (by permuting coefficients). (I now it's a bit confusing...)
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Number of classes $\pmod p$ represented by $b_1s^{n-1} + \dots + b_n$ where $ord_p(s) = n$
Now, I would like to show that these kind of evaluations reaches at least 6 classes $\pmod{11}$. For example, in the sequence (1) above one may take $\{b_1,\dots,b_5\} = \{0,0,0,0,7\}$ and it'll just generate 5 elements $\pmod {11}$, namely $7s^4, 7s^3, 7s^2, 7s, 7$. But we instead can take $\{b_1,\dots,b_5\} = \{0,0,0,7,3\}$ and it generates $3s^4 + 7s^3 \equiv 6, 3s^3 + 7s^2 \equiv 7, 3s^2 + 7s \equiv 10, 3s + 7 \equiv 8, 3 + 7s^4 \equiv 2 \pmod{11}$ (5 distinct elements). But it also generates $7s + 3 \equiv 9 \pmod{11}$, so we generate at least 6 distinct elements $\pmod{11}$. (see next..)