Let $p,q$ and $r$ be prime numbers. It is given that $p$ divides $qr − 1$, $q$ divides $rp − 1$, and $r$ divides $pq − 1$. Determine all possible values of $pqr$.

I think I'm missing something in this problem. There's no solutions online so I was wondering if you guys could show me what I'm doing wrong.

We must have one of $p,q,r = 2$ as $pq - 1$, $pr - 1$, $qr - 1$ are congruent to $0 mod 2$ if $p,q,r > 2$, so wlog let $p = 2$ (more than one of $p,q,r$ being 2 gives contradiction). Then $q$ divides $2r - 1$ (so $2r - 1 = xq$) and $r$ divides $2q - 1$ (so $2q - 1 = yr$). Solving for r we have that $(xq + 1)/2 = (2q - 1)/y$ hence $ q = (2 + y)/(4 - xy)$. This gives the only pair as $(3,5)$, so $pqr$ can only be $30$?

The way the question is phrased leads me to believe there's more to it, and also it's the last question on an olympiad so it should take longer than 5 minutes.

Re-did it, is this okay?

If we write $ rq - 1 = px $ and $ rp - 1 = qy$ we obtain $ p = \frac{r + y}{r^2 - xy} $ and $ q = \frac{r + x}{r^2 - xy} $ (and clearly we can't have $ r^2 = xy $) so $q(r + y) = p(r + x)$. Then subbing $ x = \frac{3q - 1}{p} $ and $y = \frac{3p - 1}{q}$ we obtain $r=3$ if $p$ doesn't equal $q$ (if $p=q$ then no solutions). Then it's just a matter of checking cases and we just get $(p,q,r) = (5,7,3), (2,5,3)$. But $(5,7,3)$ doesn't satisfy the fact that $r$ divides $pq - 1$ hence the only solution is $pqr = 30$. I hope that's ok

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Jul 20, 2014 at 18:21
  • $\begingroup$ Are negative primes allowed? $\endgroup$
    – Mathmo123
    Jul 20, 2014 at 18:23
  • $\begingroup$ Isn't a prime number positive by definition? Thanks Shaun I tried to make it better $\endgroup$
    – user121591
    Jul 20, 2014 at 18:25
  • 1
    $\begingroup$ I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting. $\endgroup$
    – Potato
    Jul 20, 2014 at 18:35
  • $\begingroup$ By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322 $\endgroup$
    – Potato
    Jul 20, 2014 at 18:40

1 Answer 1


The divisibility 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 $p<3$ and so $p=2$. Then $2qr\mid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover $$2qr\leq 2q+2r+qr-1,$$ or equivalently $(q-2)(r-2)\leq3$. This shows that $q=3$ and $r\in\{3,5\}$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $\{p,q,r\}=\{2,3,5\}$ and so $pqr=30$.


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