Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 11G (Number Theory)

The Fermat-Euler theorem states thatIndeed, if $r_1,\dots ,r_{\phi (n)}$ are the reduced residue classes modulo $n$, multiplication by $a$ permutes them. ThereforeThe product ${\prod }_i{r}_i$ is invertible modulo $n$, so cancellation gives the theorem.

Now let $p$ be prime. If $k\equiv 1(\mathrm{mod}p-1)$, then for $p\nmid b$, Fermat little theorem gives $b^{k-1}\equiv 1(\mathrm{mod}p)$, while $p\mid b$ makes $b^k\equiv b\equiv 0$. Thus $b^k\equiv b$ for every integer $b$.

Conversely, assume this congruence holds for every $b$. Choose a primitive root $g$ modulo $p$. Then $g^{k-1}\equiv 1(\mathrm{mod}p)$. Since $g$ has order $p-1$, one has $p-1\mid k-1$. Hence

An odd composite $N$ coprime to $b$ is a Fermat pseudoprime to base $b$ when $b^{N-1}\equiv 1(\mathrm{mod}N)$. It is a Carmichael number when this congruence holds for every $b$ coprime to $N$.

Suppose a Carmichael number has $p^a\Vert N$ with $a\ge 2$. The group $(\mathbb{Z}/p^a\mathbb{Z}{)}^{\ast }$ is cyclic, so choose a residue of orderBy the Chinese remainder theorem, extend it to an integer $b$ that is congruent to $1$ modulo every other prime-power factor of $N$. Then $\gcd (b,N)=1$, and the Carmichael congruence forces the order of $b$ modulo $p^a$ to divide $N-1$. In particular $p\mid N-1$, contradicting $p\mid N$. Therefore every Carmichael number is square-free.

Now assume $N$ is square-free. If $p-1\mid N-1$ for every $p\mid N$, then for every $b$ coprime to $N$, Fermat little theorem givesfor each $p\mid N$. The Chinese remainder theorem then gives the congruence modulo $N$, so $N$ is Carmichael.

Conversely, if $N$ is Carmichael, fix $p\mid N$. Choose a primitive root $g$ modulo $p$ and use the Chinese remainder theorem to choose $b\equiv g(\mathrm{mod}p)$ and $b\equiv 1(\mathrm{mod}N/p)$. Then $b^{N-1}\equiv 1(\mathrm{mod}p)$, so the order $p-1$ of $g$ divides $N-1$. This proves the Korselt criterion in the required form:

A Carmichael number cannot have one prime factor because it is composite. If $N=pq$ with distinct primes $p<q$, the condition $q-1\mid pq-1$ implies $q-1\mid p-1$, which is impossible. Hence every Carmichael number has at least three prime factors.

The converted source omits $r$ from the product; the PDF asks about $N=pqr$. Fix the odd prime $r$, and suppose $pqr$ is Carmichael. Its three prime factors are distinct by the Korselt criterion. Relabel $p,q$ so that $p<q$.

The divisibility conditions includeWritefor a positive integer $b$. Since $p<q$,so $1\le b<r$. Moreover,Multiplying the second divisibility by $b$ and reducing modulo $p-1$ givesThe positive integer on the right is at most $(r-1)(2r-1)$. ConsequentlyThere are only finitely many possible primes $p$ and integers $b\in \{1,\dots ,r-1\}$, and each pair determines at most one $q=(pr+b-1)/b$. Therefore