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Jan 2, 2018 at 22:50 comment added Noam D. Elkies You can't edit a comment once 5 minutes have passed, but you can still delete and replace your own comment.
Jan 2, 2018 at 21:37 answer added Aaron Meyerowitz timeline score: 1
Jan 2, 2018 at 20:44 comment added Aaron Meyerowitz Sorry, can't delete or edit earlier comment.It is true that $\sum_{a=0}^{p-1}\frac{1}{(x+a)(x+a+1)}=0$ as a formal polynomial. However there is no $x$ in $\mathbb{Z}_p$ for which it is defined. That might not matter for the question. At any rate, it seems legitimate for finite fields of order $p^e \gt p.$
Jan 2, 2018 at 20:41 comment added Aaron Meyerowitz It is true that $sum\limits_{a=0}^{p-1}\dfrac{1}{(x+a)(x+a+1)}=0\$ as a formal polynomial. However there is no $x$ in $\mathb{Z}_p$ for which it is defined. That might not matter for the question. At any rate, it seems legitimate for finite fields of order $p^e \gt p.$
Jan 2, 2018 at 20:15 comment added Aaron Meyerowitz Just to be specific, $1/(x(x+1))+1/((x+1)(x+2))+(p-2)/(x(x+2))=p/(x(x+2)).$
Jan 2, 2018 at 6:07 comment added YCor For $a_i\neq b_i$ These are $q(q-1)/2$ functions inside a subspace of dimension $q$... if $a,b,c$ are distinct, $1/((x-a)(x-b))$, $1/((x-b)(x-c))$, $1/((x-c)(x-a))$ are proportional to $1/(x-a)-1/(x-b)$, $1/(x-b)-1/(x-c)$, $1/(x-c)-1/(x-a)$ which have sum zero. So the answer is no as soon as $k,q\ge 3$. Did I miss something?
Jan 2, 2018 at 5:58 history edited user119164 CC BY-SA 3.0
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Jan 2, 2018 at 5:50 history edited user119164 CC BY-SA 3.0
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Jan 2, 2018 at 5:49 comment added abx Not always: since $\dfrac{1}{(x+a)(x+a+1)} =\dfrac{1}{(x+a)}-\dfrac{1}{(x+a+1)}$, we have $\sum\limits_{a=0}^{p-1}\dfrac{1}{(x+a)(x+a+1)}=0\ $ over any field of characteristic $p$.
Jan 2, 2018 at 5:45 comment added user119164 What if I assume I have far fewer than $p$ such functions?
Jan 2, 2018 at 5:33 review First posts
Jan 2, 2018 at 7:11
Jan 2, 2018 at 5:29 history asked user119164 CC BY-SA 3.0