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Positive Pair Poser (Posted on 2015-05-09) Difficulty: 3 of 5
Determine all possible pairs (x,y) of positive integers with gcd(4x+1, 4y-1) = 1 such that x+y divides 16xy+1

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

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution | Comment 6 of 8 |
(In reply to solution by xdog)

Bravo xdog.

My computer shows the solution set occupying all (and only)
the +ve integer lattice points on the  two lines: y = 3x – 1 and
x = 3y + 1 (tested for x, y up to 10^6) which agrees with
your conclusion.
I’ve been trying to work backwards from there, to no avail,
but your 2nd paragraph seems to be the key:

Since    4(x + y) = (4x + 1) + (4y – 1),  it follows that
x + y must divide both bracketed terms on the RHS or
neither of them. But GCD(4x + 1, 4y – 1) = 1, so
x + y cannot be a factor of both, and therefore x + y divides
neither 4x + 1 nor 4y – 1.
However, since x + y divides (4x + 1)(4x – 1) and also
(4y + 1)(4y – 1) it now follows that x + y divides both
(4x – 1) and (4y + 1), either of which allows your final
analysis.



  Posted by Harry on 2015-05-11 04:58:51

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