On average there are \( q^r +O(q^{r/2})\ \mathbb F_{q^r}\)-rational points on the the curves of genus \( g\) that are defined over \( \mathbb F_{q^r}\). What about for the curves defined over \( \mathbb F_{q}\) only? Although the average is the same if \( r\) is odd or \( r>2g\), it equals
\( q^r +q^{r/2}+O(q^{r/2}) \) otherwise.
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We prove a conjecture of Irving Kaplansky that between any two consecutive squares there is a set of distinct integers whose product is twice a squareArticle

We give various bounds for character sums in various ranges. The main idea is that if a character sum up to \( x\) is large, for a character of modulus \( q\), then it is large on \( y\)-smooth numbers, where \( y\) is of order \( \log q + (\log x)^2\), and this is the main contribution to the original character sum. We prove a weak such result assuming the Generalized Riemann Hypothesis, and unconditionally show that there are characters such that their sum up to \( (\log q)^A\) is \( \gg_A(\log q)^A\). Article

We understand the set of possible mean values of multiplicative functions whose values remain inside or on the unit circle, completely resolving this for real-valued functions. We also establish, in general, a structure theorem for large mean values, which states that the mean value must be the product of the Euler product (for the contribution of the small primes), times the solution to an integral delay equation (for the contribution of the large primes), and these can both be given explicitly. It therefore remains to better understand solutions to a certain class of integral delay equations,
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We show that if there is a Siegel zero for \( L(s,(d/.)) \) with \( d\) odd, then there are \( \sim c_d N\) prime values of the polynomial \( n^2+n+A\) (where \( d=1-4A\)) with \( n\leq N\), for all \( N\) in a wide range. There is analogous result for the polynomial \( n^2+A\) (where \( d=-4A\)) with \( d\) is even.Article

A journalistic article about Mihailescu's first breakthrough on the Catalan conjecture, showing that if \( x^p-y^q=1\) then \( p^2\) divides \( q^{p-1}-1\), and \( q^2\) divides \( p^{q-1}-1\).Article