Van Lint–MacWilliams' conjecture and maximum cliques in Cayley graphs over finite fields

S Asgarli, CH Yip - Journal of Combinatorial Theory, Series A, 2022 - Elsevier
A well-known conjecture due to van Lint and MacWilliams states that if A is a subset of F q 2
such that 0, 1∈ A,| A|= q, and a− b is a square for each a, b∈ A, then A must be the subfield …

Van Lint-MacWilliams' conjecture and maximum cliques in Cayley graphs over finite fields

S Asgarli, CH Yip - arXiv e-prints, 2021 - ui.adsabs.harvard.edu
A well-known conjecture due to van Lint and MacWilliams states that if $ A $ is a subset of
$\mathbb {F} _ {q^ 2} $ such that $0, 1\in A $, $| A|= q $, and $ ab $ is a square for each $ a …

[PDF][PDF] VAN LINT-MACWILLIAMS'CONJECTURE AND MAXIMUM CLIQUES IN CAYLEY GRAPHS OVER FINITE FIELDS

S ASGARLI, CHIHOI YIP - arXiv preprint arXiv:2106.01522, 2021 - researchgate.net
A well-known conjecture due to van Lint and MacWilliams states that if A is a subset of Fq2
such that 0, 1∈ A,| A|= q, and a− b is a square for each a, b∈ A, then A must be the subfield …

[PDF][PDF] VAN LINT–MACWILLIAMS'CONJECTURE AND MAXIMUM CLIQUES IN CAYLEY GRAPHS OVER FINITE FIELDS

S ASGARLI, CHIHOI YIP - 2022 - webpages.scu.edu
A well-known conjecture due to van Lint and MacWilliams states that if A is a subset of Fq2
such that 0, 1∈ A,| A|= q, and a− b is a square for each a, b∈ A, then A must be the subfield …

Van Lint–MacWilliams' conjecture and maximum cliques in Cayley graphs over finite fields

S Asgarli, CH Yip - 2022 - dl.acm.org
A well-known conjecture due to van Lint and MacWilliams states that if A is a subset of F q 2
such that 0, 1∈ A,| A|= q, and a− b is a square for each a, b∈ A, then A must be the subfield …

Van Lint-MacWilliams' conjecture and maximum cliques in Cayley graphs over finite fields

S Asgarli, CH Yip - arXiv preprint arXiv:2106.01522, 2021 - arxiv.org
A well-known conjecture due to van Lint and MacWilliams states that if $ A $ is a subset of
$\mathbb {F} _ {q^ 2} $ such that $0, 1\in A $, $| A|= q $, and $ ab $ is a square for each $ a …