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6 votes
2 answers
646 views

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 $N$. ...
Kurisuto Asutora's user avatar
1 vote
0 answers
96 views

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,...
Turbo's user avatar
  • 13.9k
1 vote
1 answer
552 views

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} ...
Petro Kolosov's user avatar
1 vote
1 answer
3k views

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 ...
Petro Kolosov's user avatar