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Reciprocal Equation #5 (Posted on 2013-10-15) Difficulty: 3 of 5
Find all pairs (A, B) of distinct nonzero integers with A ≠ -1 and B ≠ -1 such that (1 + 1/A) is a nonzero integer multiple of (1 + 1/B).

Prove that there are no others.

Note: (1 + 1/A) is can be a negative as well as positive integer multiple of (1 + 1/B). So, equations like: (1 + 1/A) = -2(1 + 1/B) or, (1 + 1/A) = (1 + 1/B) are permissible. Remember, A and B must be distinct.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Part of a solution | Comment 2 of 3 |
Assume A and B are both positive.  Let K be the ratio (1 + 1/A)/(1 + 1/B).  Rewrite K as [B*(A+1)]/[A*(B+1)].  A=B, K=1 is a trivial solution.

A and A+1 are coprime, so are B and B+1.  Then for K to be an integer with A!=B, a must divide B and B+1 must divide A+1.  These imply B>A and A>B simultaneously.  No value of A and B can satisfy that requirement so there are no solutions with A!=B and both A and B both positive.

Assume A and B are both negative.  Let X=1-A and Y=1-B.  X and y will both be positive.  Then K = [Y*(X-1)]/[X*(Y-1)].  This leads to the same conclusion as the previous case, the only solutions have A=B and K=1.

Assume A is negative and B is positive.  Then 0 < (1 + 1/A) < (1 + 1/B), which means K can never be an integer.  No solutions in this case.

This leaves A is positive and B is negative.  Charlie's earlier work found five nontrivial solutions.  Now to prove those are the only ones....

  Posted by Brian Smith on 2016-07-03 21:01:30
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