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.
(In reply to
Analytic Solution by Brian Smith)
If q=even and p,r=odd then in (q^2+9r)/(r+p) numerator is odd and denominator even which will never give an integer.
Likewise, r=even and p,q=odd makes (r^2+3p)/(p+q) odd/even.
Then p=2 and (q+r) evenly divides (4+2q).
2*(q+r) = 2q+2r<=4+2q gives r<=2 which is impossible.
So q+r=4+2q and r=q+4=(q+2)+2. Then the numerator of the last expression can be written (q+2)^2+4(q+2)+4+6 giving (q+2) dividing 10 evenly. Since q is an odd prime q=3.
Then the solution is (2,3,7)
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Posted by xdog
on 2023-02-04 11:58:45 |
re: Analytic Solution
