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.
First, if all of p,q,r are odd then (p^2 + 2q)/(q + r) will have an odd numerator and an even denominator, which makes it a non-integer, which violates the requirements.
So at least one of p,q,r is even, and since they are prime then at least one of p,q,r equals 2.
Case 1: p=2.
Then we have (2q+4)/(q+r), (q^2+9r)/(r+2), and (r^2+6)/(q+2) all integers.
We can say q+r >= q+2. Then (2q+4)/(q+r) <= (2q+4)/(q+2) = 2.
Then the integer requirement means that (2q+4)/(q+r) equals 1 or 2.
Subcase 1: p=2, (2q+4)/(q+r)=1
This simplifies into q=r-4. Substitute this into (r^2+6)/(q+2), which yields (r^2+6)/(r-2) = r+2 + 10/(r-2).
Now r-2 can only be 1, 2, 5, or 10, which makes r=3,4,7,12. r=4 and 12 are discarded being not prime.
Then that leaves r=3 or 7. From those q=-1 or 3, respectively. q=-1 must be discarded as being negative.
Then we have just one possible solution, (p,q,r)=(2,3,7). Verify: (p^2+2q)/(q+r)=(2^2+2*3)/(3+7)=1, (q^2+9r)/(r+p)=(3^2+9*7)/(7+2)=8, and (r^2+3p)/(p+q)=(7^2+3*2)/(3+2)=11.
Subcase 2: p=2, (2q+4)/(q+r)=2
This simplifies to r=2. Substitute this into (r^2+6)/(q+2), which yields 10/(q+2).
Now q+2 can be only 1, 2, 5, or 10, which makes q=-1, 0, 3, or 8. Only one of these is a positive prime, q=3. Then we have a possible solution (p,q,r)=(2,3,2). But this fails as (q^2+9r)/(r+2)=(3^2*9*2)/(2+2)=27/4, which is not an integer.
Case 2: q=2.
Then we have (p^2+4)/(r+2), (9r+4)/(r+p), and (r^2+3p)/(p+2).
We can say r+p >= r+2. Then (9r+4)/(r+p) <= (9r+4)/(r+2) < (9r+18)/(r+2) = 9.
Then the integer requirement means that (9r+4)/(r+p) equals 1,2,3,4,5,6,7, or 8.
Solving for r yields eight cases: 8r=p-4, 7r=2p-4, 6r=3p-4, 5r=4p-4, 4r=5p-4, 3r=6p-4, 2r=7p-4, or r=8p-4.
Cases 6r=3p-4 and 3r=6p-4 can be discarded as 4 is not a multiple of 3.
Cases 5r=4p-4 and r=8p-4 can be discarded as those require r to be a multiple of 4, which makes r not prime.
Cases 8r=p-4 and 4r=5p-4 can be discarded as those require p to be a multiple of 4, which makes p not prime.
This leaves just 7r=2p-4 and 2r=7p-4 as potential subcases.
If 7r=2p-4 we must have r=2 for r to still be prime, which makes p=9. But 9 is composite so this solution must be discarded.
If 2r=7p-4 we must have p=2 for p to still be prime, which makes r=5. But then (r^2+3p)/(p+2)=(5^2+3*2)/(2+2)=31/4 is not an integer.
So no solutions from this case.
Case 3: r=2.
Then we have (p^2+2q)/(q+2), (q^2+18)/(p+2), and (3p+4)/(p+q) all integers.
We can say p+q >= p+2. Then (3p+4)/(p+q) <= (3p+4)/(p+2) < (3p+6)/(p+2) = 3.
Then the integer requirement means that (3p+4)/(p+q) equals 1 or 2.
If (3p+4)/(p+q)=1 then we have q=2p+4, which makes q an even number larger than 2, so no solutions.
If (3p+4)/(p+q)=2 then we have p=2q-4, which we need p=2 to be a solution. Then q=3. But (q^2+18)/(p+2)=(3^2+18)/(2+2)=27/4 which is not an integer.
So after all of this we have exactly one solution (p,q,r)=(2,3,7).
Analytic Solution
