Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 1G / Solution

The Lagrange root bound over a field says that a nonzero polynomial of degree $r$ over a field has at most $r$ roots. In particular, a polynomial modular congruence of degree $r$ modulo a prime number has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive integer divisor $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative subgroup $H$ of ${\mathbb{F}}_p^{\times }$. By the fact that every finite multiplicative subgroup of a field is cyclic, $H=⟨g⟩$ is a cyclic group of order $d$. The solutions in $H$ of $x^e=1$ areso there are at least $e$ of them. The Lagrange root bound over a field gives at most $e$ roots in all of ${\mathbb{F}}_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the Chinese remainder theorem for unit groups, a base $b$ is a Fermat-pseudoprime base precisely when its two components satisfyThe power roots in a finite field andshow that each component has ten choices. Therefore there areFermat-pseudoprime bases.

To impose the strong pseudoprime condition, write $n-1=2^{s}m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is

Solved by gpt-5.6-sol high.