200 Followers Problem
p 1 + q 1 + p q 1 = n 1
Over all natural numbers n , how many ordered pairs of primes ( p , q ) satisfy the equation above?
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1 solution
Ah I forgot the permutation.
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Same here :(
Yes , same method. I think I have solved this question before in some book (I dont remember).
Wouldn't it be better if we used SFFT? BTW, nice solution!
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I don't think SFFT will help here. How would you use it?
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Can't SFFT be used as follows ??
p 1 + q 1 + p q 1 = n 1
Multiplying by (p * q * n) on both sides ⇒ n q + n p + n = p q ⇒ p q − n p − n q = n ⇒ p q − n p − n q + n 2 = n + n 2 ⇒ ( 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
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)
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@Mohit Sharma – Why must we have p − n = n , q − n = n + 1 ? Why can't we have (say) p − n = 2 n , q − n = 2 ( n + 1 ) ?
Because you now have much less control over what 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.
p 1 + q 1 + p q 1 = n 1
or, p q p + q + 1 = n 1
or, n ( p + q + 1 ) = p q
As p , q are primes so four cases are here
Case 1
n = 1 & p + q + 1 = p q
Case 2
n = p & p + q + 1 = q
Then p = − 1
Which is not possible.
Case 3
n = p q & p + q + 1 = 1
Then p + q = 0
Which is not possible.
Case 4
n = q & p + q + 1 = p
Then q = − 1
which is not possible.
So only Case 1 is possible.
Taking it we will get 2 ordered pairs ( 2 , 3 ) , ( 3 , 2 )