Carmichael Number: Proving $(d+p)(p-1)$ Divisor of $q-1$

[peteryellow]

$n=pqr$ is a Carmichael number. If $q-1|pr-1$ and $r-1|pq-1$ then show that q-1 is a divisor in
$(d+p)(p-1).$

 

[peteryellow]

ok here it is the clear version of it.

n=pqr is carmicahelnumber. if we have that r-1|pq-1 then I have shown that
pq-1=d(r-1) where d is [2;p-1]. Moreover I have shown that q-1|d(r-1)-p+1.
Now I want to show that if q-1|pr-1 is also fulfilled then q-1 is divisor in (d+p)(p-1). Do you understand it?

 

[rscosa]

Hi!
I suggest you to read more before publish your problems since this is a very well known question (Wikipedia).