Question

The number of pairs of prime numbers (p,q) satisfying the condition 51/100<1/p + 1/q< 5/6 will be ​

Answer 1

Answer:

There is only one solution p=q=2 & r=3 , we will prove it for a positive integer n

We deduce that , p(p+1)=n(n+1)−q(q+1)=(n−q)(n+q+1) and we must have n>q

Since p is a prime we must have p|(n−q) or p|(n+q+1) , Now if p|(n−q) then p≤(n−q) which implies p(p+1)≤(n−q)(n−q+1)⟹(n−q)(n+q+1)≤(n−q)(n−q+1) and therefore (n+q+1)≤(n−q+1) which is impossible and thus p|(n+q+1) .

For some positive integer k we have n+q+1=kp & thus putting it into original equation we yeild p(p+1)=kp(n−q) which is p+1=k(n−q) , For k=1 we have p+q+1=n & n+q+1=p which yields p−n=n−p on subtracting which is absurd and thus k>1 .

Note that kp−1=n+q,p=k(n−q)−1 & so we have the identity 2q=(n+q)−(n−q)=kp−1−(n−q)=k(k(n−q)−1)−(n−q)=(k+1)[(k−1)(n−q)−1] , With these and the condition that k≥2⟹k+1≥3 .

2q has divisors 1,2,q,2q only which implies k+1=q or k+1=2q only.

If k+1=q , (k−1)(n−q)=3 hence (q−2)(n−q)=3 which akes either q−2=1 & n−q=3 which yields the solutions (p,q,k)=(5,3,2) and thus n=6 .

Else if q−2=3 & n−q=1 which makes p=3,q=5,n=6 .

Now if k+1=2q then (k−1)(n−k)=2 and 2(q−1)(n−q)=2 which leads to q−1=1 & n−q=1 and thus p=2,q=2,n=3

Thus we have the following solutions in positive integer n & primes p,q which are (p,q,n)=(5,3,6),(3,5,6),(2,2,3) and it is clear that only one prime solution exist p=q=2,r=3 .

Hope it helps !