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 Divide The Cubic, Get The Pairs (Posted on 2007-05-26)
Find all possible positive integers (p,q) with p≤q such that (q³+1)/(pq-1) is an integer.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 Solution | Comment 1 of 5

Let f = (q^3+1)/(pq-1) [f>0].  Then rewrite the equation as q^3+1 = (pq-1)*f.

If q=2 then q^3+1 is odd and then both factors must be odd.

If q>=3 then f = -1 (mod q).  This is due to congruence mod q of the equation which yields 1 = (-1)*f (mod q).

Let f = nq-1 [n>0].  This satisfies the congruence above for q>=3 and satisfies being odd for q=2.

Then q^3+1 = (pq-1)*(nq-1).  Rearranging and dividing each side by q yields: q^2 - p*n*q + p + n = 0.

Applying the quadratic fomula yields q = (p*n +/- sqrt[(p*n)^2 - 4*(p+n)])/2

q will be an integer when (p*n)^2 - 4*(p+n) is a perfect square.  (p*n+1)>(p+n) for p,n >= 2.

If p*n >= 7, both of the following inequalities are true:
(p*n)^2 > (p*n-2)^2 - 8 > (p*n-3)^2
(p*n)^2 > (p*n)^2 - 4*(p+n) > (p*n)^2 - 4*(p*n+1) = (p*n-2)^2 - 8

Combining these inequalites yields: (p*n)^2 > (p*n)^2 - 4*(p+n) > (p*n-3)^2  So, if p*n >= 7, then the only perfect square values of (p*n)^2 - 4*(p+n) are (p*n-1)^2 and (p*n-2)^2.

If (p*n)^2 - 4*(p+n) = (p*n-1)^2 then 4p + 4n = 2pn - 1, which is false for all integers.
If (p*n)^2 - 4*(p+n) = (p*n-2)^2 then 0 = 4, which is false.
Therefore p*n must be less than 7.

The pairs (p,n) with 1<=p<=n<=6 which (p*n)^2 - 4*(p+n) is a square are: (1,5), (2,2), (2,3).

These pairs generate nine solutions (p,q) = (1,2), (1,3), (5,2), (5,3), (2,2), (2,1), (2,5), (3,1), (3,5).

Edited on May 27, 2007, 12:57 am
 Posted by Brian Smith on 2007-05-27 00:56:17

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