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 Diophantine Pair Decision (Posted on 2016-08-13)
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?

 No Solution Yet Submitted by K Sengupta No Rating

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 Dead Ends? | Comment 1 of 3
Well, X or Y can be 1.  This gives rise to 3 solutions (1,1), (1,11) and (11,1).

(Y^2 - 12)/X is an integer, so if X is divisible by 2, then so is Y.
Similarly, if X is divisible by 3 then so is Y.
But neither can be the case, because X and Y are relatively prime.
So X and Y cannot be divisible by 2 or 3.

If (Y^2 - 12)/X and (X^2 - 12)/Y are both integers, then so are their products.
In other words, (X^2*Y^2 -12X^2 - 12Y^2 + 144)/XY is an integer.
Then, subtract XY to get (-12X^2 - 12Y^2 + 144)/XY is an integer.
X and Y are not divisible by 2 or 3, so we can divide by -12 to get
(X^2 + Y^2 - 12)/XY is an integer.
Add 2 to get ((X+Y)^2 - 12)/XY is an integer
Or, subtract 2 to get ((X-Y)^2 - 12)/XY.

But none of that seems to help.

Also not obviously helpful:  Since X and Y are odd, let X = 2A + 1, Y = 2B + 1.
That seems unfruitful, especially since A and B do not need to be relatively prime.

Out of ideas.

 Posted by Steve Herman on 2016-08-13 20:03:47

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