1. Sep 26, 2007

### menager31

Find all triples of primes (p,q,r), that pq+qr+rp and p^3+q^3+r^3−2pqr are divisible by p+q+r.
I really don't know how to start, (of course i've been trying)

2. Sep 26, 2007

### AlephZero

Hint: start by multiplying out (p+q+r)(p2 + q2 + r2)

3. Sep 26, 2007

### HallsofIvy

By "p2+ q2+ r2" do you mean p^2+ q^2+ r^2 ?

4. Sep 26, 2007

### menager31

it doesn't simplify those quotients( I was counting about an hour)

5. Sep 26, 2007

### menager31

6. Sep 26, 2007

### AlephZero

$(p+q+r)(p^2 + q^2 + r^2)$
$= p^3 + q^3 + r^3 + pq(p+q) + qr(q+r) + rp(r+p)$
$= p^3 + q^3 + r^3 + (p+q+r)(pq+qr+rp) - 3pqr$
$= (p^3 + q^3 + r^3 - 2pqr) + (p+q+r)(pq+qr+rp) - pqr$

The whole expression is divisible by (p+q+r)
If the given conditions hold, pqr is divisible by p+q+r.

But p,q,r are primes, therefore...

Notes: the question doesn't say p,q,r are distinct.
And so far, we haven't used the fact that p+q+r divides pq +qr + rp.

7. Sep 27, 2007

### menager31

ok, now i see your solution, thanks, big thanks

8. Nov 4, 2007