perplexus.info :: Just Math : Reciprocal Equation #5

Reciprocal Equation #5 (Posted on 2013-10-15)

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.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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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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