Thomas Bloom

Thomas F. Bloom is a mathematician, who is a Royal Society University Research Fellow at the University of Manchester.^[1] He works in arithmetic combinatorics and analytic number theory.

Thomas Bloom
Nationality	British
Alma mater	University of Oxford University of Bristol
Awards	Royal Society University Research Fellowship
Scientific career
Institutions	University of Cambridge University of Oxford University of Bristol University of Manchester
Doctoral advisor	Trevor Wooley
Other academic advisors	Timothy Gowers

Education and career

Thomas did his undergraduate degree in Mathematics and Philosophy at Merton College, Oxford. He then went on to do his PhD in mathematics at the University of Bristol under the supervision of Trevor Wooley. After finishing his PhD, he was a Heilbronn Research Fellow at the University of Bristol. In 2018, he became a postdoctoral research fellow at the University of Cambridge with Timothy Gowers. In 2021, he joined the University of Oxford as a Research Fellow.^[2] Then, in 2024, he moved to the University of Manchester, where he also took on a Research Fellow position.

Research

In July 2020, Bloom and Sisask^[3] proved that any set such that ${\displaystyle \sum _{n\in A}{\frac {1}{n}}}$ diverges must contain arithmetic progressions of length 3. This is the first non-trivial case of a conjecture of Erdős postulating that any such set must in fact contain arbitrarily long arithmetic progressions.^[4][5]

In November 2020, in joint work with James Maynard,^[6] he improved the best-known bound for square-difference-free sets, showing that a set ${\displaystyle A\subset [N]}$ with no square difference has size at most ${\displaystyle {\frac {N}{(\log N)^{c\log \log \log N}}}}$ for some ${\displaystyle c>0}$.

In December 2021, he proved ^[7] that any set ${\displaystyle A\subset \mathbb {N} }$ of positive upper density contains a finite ${\displaystyle S\subset A}$ such that ${\displaystyle \sum _{n\in S}{\frac {1}{n}}=1}$.^[8] This answered a question of Erdős and Graham.^[9]