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Unequally Yoked III (Posted on 2009-09-01) Difficulty: 3 of 5
Determine all possible nonzero integer(s) P and Q, with PQ, that satisfy this equation.

                                  PP+2Q = QQ+2P

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution analytical solution, completed (spoiler) Comment 3 of 3 |


Let P = x^k, Q = x^m,  k < m

Per my previous post, it follows that

  x^(m-k) = (2m-k)/(2k-m) 
  where x, m and k are positive integers and k < m < 2k

Let a = m - k
  Then 
  x^a = (2a + k)/(k - a) = 1 + 3a/(k-a)
  where x, a and k are positive integers and a < k  and x > 1
  

If a = 1, then

x = 1 + 3/(k-1) has two integral solutions:

 a) k = 2, x = 4, m = k+a = k+1 = 3, P = x^k = 16, Q = x^m = 64
 b) k = 4, x = 2, m = k+a = k+1 = 5, P = x^k = 16, Q = x^m = 32
  

If a = 2, then

x^2 = 1 + 6/(k-2) has one integral solution:

 c) k = 4, x = 2, m = k+a = k+2 = 6, P = x^k = 16, Q = x^m = 64
 
If a = 3, then

x^3 = 1 + 9/(k-3) has no integral solutions.
    the rhs is integral only if k = 4, 6, or 12
    but then it equals 10, 4, or 2, none of which are cubes
    
And there are no integral solutions if a > 4, because

   1 + 3a/(k-a) <= 3a + 1
   and x^a >= 2^a > 3a
   
   so x^a <> 1 + 3a/(k-a)
    
So the only solutions where P < Q are (P,Q) = (16,32) and (16,64)

And of course, the other solutions to the problem are  (P,Q) = (32,16) and (64,16)

Edited on September 4, 2009, 2:08 pm
  Posted by Steve Herman on 2009-09-04 03:48:03

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