20.36 (2022)

The free Burnside group of exponent four on three generators, $B(3, 4)$, has order $2^{69}$ as shown by Bayes, Kautsky, and Wamsley (1974). Their proof is based on a theorem of Sanov (1940) which shows that $B(n, 4)$ is finite. Sanov’s proof for $B(3, 4)$ uses more than $2^{32}$ fourth powers, because the subgroup of $B(3, 4)$ generated by two of its generators and the square of the third has order $2^{32}$. It is also known that $B(3, 4)$ needs at least 105 relations to define it, as shown by Havas and Newman (1980).
$\qquad$ a) Can $B(3, 4)$ be defined with fewer than a million fourth powers?
$\qquad$ b) Can $B(3, 4)$ be defined with fewer than a thousand fourth powers?
$\qquad$ c) What is the smallest number of fourth powers which define $B(3, 4)$?