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Problem Of The Week 18

with 2 comments

Let p,q be prime numbers and n be a natural numbers. Find the number of ordered triplets (p,q,n) such that


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


Written by Implex

September 18, 2008 at 3:03 pm

2 Responses

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  1. 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.

    Celebrating Life

    September 18, 2008 at 3:21 pm

  2. yupp n=pq/(p+q+1)

    p+q+1 divides pq

    so this can happen only when p+q+1=pq
    p=2,3 q=3,2

    so two triplets !


    September 18, 2008 at 3:47 pm

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