14 votes
Accepted

Number of Laurent monomials of n variables with degree at most d

One such formula is $$\sum_{p=0}^n \binom{n}{p} \binom{d}{p} \binom{d+n-p}{n-p}.$$ To derive this, let $P \subseteq [n]$ be the set of variables with positive exponents and let $p = |P|$. There are $\...
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12 votes
Accepted

Multiple roots of polynomials with coefficients $\pm 1$

Question P. Can a polynomial $P(x)=\sum_{n=0}^ma_nx^n$ with coefficients $a_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$? Yes. The following four Littlewood ...
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  • 4,257
10 votes

Solve this sextic

The RHS only depends on $n(n+1)$, specifically it can be written as $$ 4a\left(8(n(n+1))^3+6(n(n+1))^2+n(n+1)+\frac12\right)+4bn(n+1)+2c, $$ so you have to use the standard formulas to solve a cubic ...
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9 votes
Accepted

Divergence of primes dividing polynomials

Yes, the series diverges. We can reduce easily to the case of irreducible monic $Q$. Next, let $\alpha_Q(p)$ be the number of roots of $Q(x)$ in $\mathbb{Z}/p\mathbb{Z}$. Note that $M_Q$ is the set of ...
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6 votes

Multiple roots of polynomials with coefficients $\pm 1$

Not an answer, only some sort of evidence for A, hence community wiki. There seems to be an algorithm that produces a sequence of polynomials $(P_n)$ like in P, with $P_0=1$, $P_{n+1}=P_n\pm x^{n+1}$, ...
6 votes

Solve this sextic

Mathematica finds a closed-form expression for the solutions $n$ as a function of $m$ of the equation $$m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c.$$ The expressions for general $a,b,...
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5 votes

MacMahon Master Theorem for non-matching coefficients

The reason I asked this question is that I found such a generalization of MMT and didn't know if it exists in the literature. The proof makes extensive use of operator calculus: the differential ...
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4 votes
Accepted

Checking presence of a specific term in product polynomial

You can solve the problem via integer linear programming as follows. Let $a_{ijk}$ be the power of $x_j$ in term $k$ of $p_i$, that is, $p_i = \sum_k \prod_j x_j^{a_{ijk}}$. Let binary decision ...
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  • 3,760
4 votes

Do polynomial values rarely have large multiple prime factors?

The question of showing that a multivariable polynomial takes on many squarefree values, or more generally controlling the size of the largest square factor, is referred to in the literature as the &...
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4 votes

Invertible linear transformation on the space of polynomials

Let $\Delta_x:\ f(x)\rightarrow f(x+1)-f(x)$ be the forward difference operator then $\mathscr{A}=2I+\Delta_x$ where $I$ is the identity operator. Using formal expansion of operators $$\mathscr{A}^{-1}...
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  • 1,445
3 votes

Alon-Füredi for homogeneous polynomials

If $0\notin B$, then $$F(x,y) = y^n \left(c_n \left(\frac{x}{y}\right)^n+\cdots+c_0\right).$$ Let $S=\{x/y:\ x\in A,y\in B\}.$ For $F$ to vanish on all the points of $A\times B$ except $1$, we must ...
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  • 10.7k
3 votes
Accepted

Alon-Füredi for homogeneous polynomials

In general homogeneity does not improve the bound. Take $A=\{1,q,q^2,\ldots,q^{a-1}\}$, $B=\{1,q,q^2,\ldots,q^{b-1}\}$, $f(x,y)=\prod_{i=-(b-1)}^{a-2}(x-q^iy)$.
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  • 88.5k
2 votes

A question about generalized harmonic numbers modulo $p$

Glaisher's I-numbers are described in J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
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  • 13.3k
1 vote
Accepted

Changing base field for sum of polynomials

Not always. Take $k=3$, $n=2$, $f_1=\sqrt{2}x_1^2+x_2^2$, $f_2=x_3^2$.
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  • 88.5k
1 vote

Irrational combination of rationally independent polynomials

Yes: if $p$ has rational coefficients, then (with fixed $p$) the real coefficients $t_j$'s are uniquely determined, since the polynomials $p_j$ are linearly independent (this property does not depend ...
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  • 88.5k
1 vote

Why the sequence of Bernstein polynomials of $\sqrt x$ is increasing?

We have $$B_n(f,\frac{u}{u+v}) = \frac{1}{(u+v)^n} \sum_{k=0}^n \binom{n}{k} f(\frac{k}{n})\,u^k v^{n-k}$$ so $$B_n(f,\frac{u}{u+v})- B_{n+1}(f,\frac{u}{u+v})=\\= \frac{1}{(u+v)^{n+1}}\left((u+v)\sum_{...
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