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Diophantine Pair Decision (Posted on 2016-08-13) Difficulty: 3 of 5
Each of X and Y is a distinct positive integer, such that:
  1. X and Y are relatively prime, and:
  2. Each of (X2 – 12)/Y and (Y2 – 12)/X is an integer.
Does there exist an infinite numbers of pairs (X,Y) satisfying the given conditions?

Give reasons for your answer.

No Solution Yet Submitted by K Sengupta    
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Hints/Tips OEIS to save the day! | Comment 2 of 3 |
I hit the same dead ends that Steve did, so I decided use a trick Charlie is fond of - write a program to brute force some small values.  That found one new ordered pair: (11,109).

With this third solution sharing 11 with the prior solution (1,11) I suspected that there are indeed an infinite number of solutions.  Specifically a recursive sequence starting 1,11,109.  11,109 is an unusual pair of consecutive terms so I tried my luck at the OEIS.

The first hit is sequence A054320 which continues as 1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, .......  This has the recursive generator t(n+2) = 10*t(n+1)-t(n).  Because the first two terms are coprime then every pair of consecutive terms is coprime.

Plug in any consecutive pair for X,Y.  Then the two integer results are the term before and the term after the pair.  For example, take the (11,109) pair: (11^2-12)/109 = 1 aka the preceding term, and (109^2-12)/11 = 1079 aka the following term.

I have not formally proven this, but this comment in the OEIS entry might help: "For all members x of the sequence, 6*x^2 + 3 is a square."

  Posted by Brian Smith on 2016-08-13 21:44:57
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