User tim browning - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-19T05:51:50Zhttp://mathoverflow.net/feeds/user/30668http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/125224/rational-points-on-a-sphere-in-mathbbrd/125750#125750Answer by Tim Browning for Rational points on a sphere in $\mathbb{R}^d$Tim Browning2013-03-27T17:56:12Z2013-03-27T17:56:12Z<p>With regards to Q1 you can get a finer notion of density by counting the number $N(B)$, say, of rational points with height at most $B$, and consider how this quantity behaves as $B\rightarrow \infty$. I would think of this problem projectively. In this case rational points would be $[x_1,\dots,x_n,y]$, with $x_1,\dots,x_n,y$ integers such that $\gcd(x_1,\dots,x_n,y)=1$, for which $$
x_1^2+\cdots +x_d^2=y^2.$$
One could then define the height of a point to be $\max(|x_i|,|y|)$. Using the Hardy-Littlewood circle method, for example, one can prove (for $d\geq 3$) that
$$
N(B)\sim B^{d-1}\sigma_\infty\prod_p\sigma_p,
$$
as $B\rightarrow \infty$, where $\sigma_v$ is the density of points on the quadric in the completion $\mathbb{Q}_v$. </p>
http://mathoverflow.net/questions/45511/density-of-values-of-polynomials-in-two-variables/119288#119288Answer by Tim Browning for Density of values of polynomials in two variablesTim Browning2013-01-18T20:35:23Z2013-01-18T20:35:23Z<p>I would certainly expect $g(N)$ to be rather small when $f$ has larger degree. Let us consider the special case where $f(x,y)=x^d+y^d$, for $d\geq 3$. Consider
the arithmetic function $r(n)$, which counts the number of $(x,y)\in \mathbb{N}^2$ such that $n=f(x,y)$. The first moment of $r(n)$ is easily understood via the geometry of numbers. The second moment was looked at by Hooley (On another sieve method and the numbers that are a sum of two $h$th powers. <em>Proc. London Math. Soc.</em> <strong>43</strong> (1981), 73-109).</p>
<p>As a consequence of this there exists an explicit constant $c>0$ such that there are asymptotically $c N^{2/d}$ integers $n\leq N$ which can be written as $x^d+y^d$, and furthermore, almost all of these have essentially just one representation.</p>
http://mathoverflow.net/questions/69493/integral-roots-to-degree-d-forms-in-four-variables-inside-a-box/119026#119026Answer by Tim Browning for Integral roots to degree $d$-forms in four variables inside a boxTim Browning2013-01-15T21:03:04Z2013-01-15T21:03:04Z<p>There is a result of the form you want in the literature. Take a look at Theorem 4 in Heath-Brown's paper <em>The density of rational points on curves and surfaces</em> (Annals of Math., <strong>155</strong> (2002), 553–595). An easy generalisation of this gives the following. Let $G\in \mathbb{Z}[X_0,\ldots, X_n]$ be a primitive form of degree $d\geq 2$, defining a projective hypersurface $Z\subset \mathbb{P}^n$. Then either
$$
\|G\|\ll_{d,n} B^{d\binom{d+n}{n}},
$$
or else there exists a form $G′\in \mathbb{Z}[X_0,\ldots, X_n]$ of degree $d$, which is not proportional to $G$, such that $G′$ vanishes at each point $x \in Z\cap \mathbb{P}^n(\mathbb{Q})$ which has height $H(x) \leq B$. </p>
<p>Upshot: In your case, either $\|F\|$ is small or the points you are interested actually lie on a curve in $\mathbb{P}^3$. In the latter case you get the bound $N(F,B)=O_{d}(B)$ unless the curve you produce contains a line (which must then lie in the surface $F=0$).</p>
http://mathoverflow.net/questions/14076/irreducibility-of-polynomials-in-two-variables/118956#118956Answer by Tim Browning for Irreducibility of polynomials in two variablesTim Browning2013-01-15T10:06:09Z2013-01-15T10:06:09Z<p>Quite a useful result can be found in Schmidt's lecture notes on <em>Equations over finite fields</em> (Theorem III.1B in SLNM 536). Suppose that $K$ is any field and let $f(x,y)=c_0y^d+c_1(x)y^{d-1}+
\cdots+c_d(x) \in K[x,y]$, with $c_0 \neq 0$. Let
$$
\psi(f)=\sup_{1\leq
i\leq d}\frac{\deg c_i}{i}.
$$
Then $f$ is absolutely irreducible over $K$ provided that that $\psi(f)=m/d$ with $\gcd(m,d)=1$.</p>
<p>This shows, for example, that the polynomial $f(x,y)=g(x)-h(y)$ is irreducible when $\deg g$ and $\deg h$ are coprime.</p>