200 Followers Problem

1 p + 1 q + 1 p q = 1 n \large\frac{1}{p}+\frac{1}{q}+\frac{1}{pq}=\frac{1}{n}

Over all natural numbers n n , how many ordered pairs of primes ( p , q ) (p,q) satisfy the equation above?

3 2 Infinitely Many 4 1 9

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1 solution

Kalpok Guha
Jun 17, 2015

1 p + 1 q + 1 p q = 1 n \frac{1}{p}+\frac{1}{q}+\frac{1}{pq}=\frac{1}{n}

or, p + q + 1 p q = 1 n \frac{p+q+1}{pq}=\frac{1}{n}

or, n ( p + q + 1 ) = p q n(p+q+1)=pq

As p , q p,q are primes so four cases are here

Case 1

n = 1 n=1 & p + q + 1 = p q p+q+1=pq

Case 2

n = p n=p & p + q + 1 = q p+q+1=q

Then p = 1 p=-1

Which is not possible.

Case 3

n = p q n=pq & p + q + 1 = 1 p+q+1=1

Then p + q = 0 p+q=0

Which is not possible.

Case 4

n = q n=q & p + q + 1 = p p+q+1=p

Then q = 1 q=-1

which is not possible.

So only Case 1 is possible.

Taking it we will get 2 \boxed{2} ordered pairs ( 2 , 3 ) , ( 3 , 2 ) (2,3),(3,2)

Ah I forgot the permutation.

Jaka Ong - 5 years, 11 months ago

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Same here :(

Dylan Pentland - 5 years, 11 months ago

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Me too :))

Kazem Sepehrinia - 5 years, 11 months ago

Yes , same method. I think I have solved this question before in some book (I dont remember).

Nihar Mahajan - 5 years, 12 months ago

Wouldn't it be better if we used SFFT? BTW, nice solution!

B.S.Bharath Sai Guhan - 5 years, 12 months ago

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I don't think SFFT will help here. How would you use it?

Calvin Lin Staff - 5 years, 11 months ago

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Can't SFFT be used as follows ??

1 p + 1 q + 1 p q = 1 n \frac{1}{p} + \frac{1}{q} + \frac{1}{ pq} = \frac{1}{n}

Multiplying by (p * q * n) on both sides n q + n p + n = p q \Rightarrow nq + np + n = pq p q n p n q = n \Rightarrow pq-np-nq = n p q n p n q + n 2 = n + n 2 \Rightarrow pq-np-nq+n^{2} = n + n^{2} ( p n ) ( q n ) = n ( n + 1 ) \Rightarrow (p-n) (q-n) = n(n+1)

Now, there are two possibilities: p = 2 n , q = 2 n + 1 ; p = 2 n + 1 , q = 2 n p = 2n, q = 2n + 1 ; p = 2n+1, q = 2n

Now, only value of n for which p and q are both prime is n = 1, as for any other value of n, one of p or q would be a multiple of 2.

Using n=1 gives 2 ordered pairs : (2,3), (3,2) \text{Using n=1 gives } \boxed{2} \text{ordered pairs : (2,3), (3,2)}

Mohit Sharma - 5 years, 11 months ago

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@Mohit Sharma Why must we have p n = n , q n = n + 1 p - n = n, q - n = n+1 ? Why can't we have (say) p n = n 2 p-n = \frac{ n}{2} , q n = 2 ( n + 1 ) q -n = 2 (n+1) ?

Because you now have much less control over what p n , q n p-n, q-n are (IE need not be prime), the unique factorization will no longer work.

That is why I do not believe that SFFT will be helpful.

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin Thank you for the explanation.

Mohit Sharma - 5 years, 11 months ago

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