Find all possible triplet(s) (p, q, r) of
prime numbers such that, each of:
(p2 + 2q)/(q + r), (q2 + 9r)/(r + p), and
(r2 + 3p)/(p + q)
is a positive integer.
Prove that no further triplet is possible in conformity with the given conditions.
Note: Adapted from a problem appearing in a shortlist of Junior Balkan Mathematical Olympiad.
*** Computer program assisted solutions are welcome, but a semi-analytic methodology, that is: hand calculator and p&p, is preferred.
Consider (p^2 + 2q)/(q + r). If p, q and r are all odd, then this is an odd number divided by an even number, which cannot be an integer. Therefore, at least one of p, q and r must equal 2.
They cannot all equal to 2, because that would make (r^2 + 3p)/(p + q) = 10/4
Can exactly two of them = 2?
If it is q & r, then (p^2 + 2q)/(q + r) = (p^2 + 4)/4, which is not an integer for an odd p.
If it is q & p, then (r^2 + 3p)/(p + q) = (r^2 + 6)/4, which is not an integer for an odd r.
If it is p & r, then (q^2 + 9r)/(r + p) = (q^2 + 18)/4 which is not an integer for an odd q.
Therefore, exactly one of them equals 2, and the other two are odd primes.
An analytical start
