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.
pr=primes(nthprime(900));
for pp=1:900
for qq=1:900
for rr=1:900
p=pr(pp);q=pr(qq);r=pr(rr);
a=(p^2+2*q)/(q+r);
if a==round(a)
b=(q^2+9*r)/(r+p);
if b==round(b)
c=(r^2+3*p)/(p+q);
if c==round(c)
disp([p,q,r])
disp([a b c])
end
end
end
end
end
end
nthprime(900)
finds only
2 3 7 as the primes
1 8 11 as the three fractions (resulting integers)
and reports having tried up to the 900th prime (6997) for each of p, q and r.
|
|
Posted by Charlie
on 2023-02-04 09:54:58 |
computer findings
