Smd Qr-1

0 views

Skip to first unread message

Reggie Lamborn

unread,

Aug 5, 2024, 1:00:24 AM8/5/24

to resanlere

Thedivisibility criteria ensure that $pqr$ divides$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$and hence that $pqr\mid pq+pr+qr-1$. Without loss of generality $p\leq q\leq r$ and so$$3qr-1\geq pq+pr+qr-1\geq pqr,$$which shows that $p