All Questions
4 questions
6
votes
2
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646
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Number of solutions of $am \equiv bn \pmod{q}$
Let $q$ be a (large) prime. Let $N$ be a positive integer of size${}\approx \sqrt{q}$. Let $\mathcal{M}$ be an arbitrary subset of $\{1, \dots, q\},$ such that $\mathcal{M}$ has cardinality $N$. ...
1
vote
0
answers
96
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Polynomial composition utilizing polynomials in two different finite fields
At every $n\in\mathbb N$ (all polynomials are of degree $O(1)$) is there $g_{3,1}^{(n)},\dots,g_{3,k}^{(n)}\in\mathbb F_3[x_1,\dots,x_n]$ at $k=\mathsf{poly}(n)$ and $g_2^{(n)}\in\mathbb F_2[x_1,\dots,...
1
vote
1
answer
552
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Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$
Review the main result of mathoverflow.net/questions/297900, that is the identity
\begin{equation}\label{f1}
n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j,
\end{equation}
...
1
vote
1
answer
3k
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Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}, \ m=1,2,....$
Consider a sum $$\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j$$
which returns an odd power $n^{2m+1}$ of $n$, for $\ m=0,1,2,...$ given fixed $A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$. The ...