Let p,q be prime numbers and n be a natural numbers. Find the number of ordered triplets (p,q,n) such that
1/p+1/q+1/(pq)=1/n
A) 0 B) 1 C) 2 D) 4 E) none of These
Tipster: All prime( >3) squares are of the form 6k+1 where k is a positive integer. Note this is a necessary condition not a sufficent one
I think there is only one possibility(p,q)=(2,3)..The given equation reduces to (p+q+1)/p*q..Assume p>Q therefore Q must be a factor of p+q+1 and the other factor must divide p, but p is prime and hence cannot have any factors other than 1 and itself.But for (2,3) p=q+1 but no other pair of primes has this luxury.
ReplyDeleteyupp n=pq/(p+q+1)
ReplyDeletep+q+1 divides pq
p+q+1>p,q,1
so this can happen only when p+q+1=pq
p=2,3 q=3,2
so two triplets !