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QUESTION
Let $n=q_1q_2\ldots q_k$, where the $q_i$ are distinct primes and
$k>1$. Suppose that for each $i,\ q_i-1|n-1$. Show that $n$ is a
Carmichael number.
Hence find a Carmichael number of the form $7.23.q$ where $q$ is
an odd prime.
ANSWER
$n=q_1q_2\ldots q_k,\ q_i$ distinct primes, and $k>1$. Thus, as
$k>1,\ n$ is composite.
Suppose gcd$(b,n)=1$. Thus gcd$b,q_i)=1$ for each $i$. Hence
$b^{p_i-1}\equiv1$ mod $q_i$ by Fermat's Little Theorem. But
$q_i-1|n-1$, say $(q_i-1)s=(n-1)$ for some $s$. Thus
$b^{n-1}=(b^{q_i-1})^s\equiv1^1\equiv1$ mod $q_i$. Thus
$q_i|b^{n-1}-1$, and this is true for each $i$, so we get
$q_1q_1\ldots q_k|b^{n-1}-1$ by cor.1.7. But$n=q_1q_2\ldots q_k$,
so $n|b^{n-1}-1$ and $b^{n-1}\equiv1$ mod $n$. Thus $n$ is a
Carmichael number as required.
Suppose $q$ is an odd prime $\neq 7,23$. By the above proof,
$n=7.23.q$ will be a Carmichael number if each of 6,22,$q-1$
divides $n-1$.
Consider the equation $n=7.23.q$ modulo 6. Since $6|n-1$, we have
$n\equiv1$ mod 6, and so $\equiv1.-1.q$ mod 6, giving $q\equiv -1$
mod 6. Similarly, reducing the equation $n=7.23.q$ mod 22, we get
$1\equiv7.1.q$ mod 22, or, on multiplying by 3, $3\equiv-1.q$ mod
22. Thus $q\equiv -3$ mod 22. Finally, reducing the equation
$n=7.23.q$ mod $(q-1)$, we get $1\equiv 7.23.1$ mod $q-1$, ei.e.
$160\equiv 0$ mod $(q-1)$, so that $q-1$ divides 160. To find a
prime satisfying all three requirements, we may start listing
positive integers congruent to $-3$ mod 22, checking each in turn
to see of they satisfy the other two requirements, $q\equiv -1$
mod 6, and $q-1$ divides 160. Our process either produces a
positive integer satisfying our requirements, or leads to numbers
larger than 160, so that we could conclude that no such integer
$q$ existed. In any case the process will terminate. We check
$q=19$m which fails, then $q=41$, which is prime and satisfies all
our conditions, so $7.23.41$ is a suitable Carmichael number.
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