Determine all possible pairs (x,y) of positive integers with gcd(4x+1, 4y1) = 1 such that x+y divides 16xy+1
I took as my starting point Steve's observation that if (x+y) divides 16xy+1, it also divides (4y+1)(4y1) and likewise divides (4y+1)(4y1).
So (x+y) will divide the product of those terms and by the problem's gcd restriction it will divide (4x1)(4y+1).
But that product = 16xy+4x4y1 = 16xy+4x+4y8y1 = (16xy+1)+(4x+4y)(8y+2).
Then (x+y) factors (8y+2) and since (x+y) is oddit equals 1mod16dividing by 2 is OK which gives (x+y)*k = (4y+1) for integer k.
Now we're down to cases. k=4 is already too big. k=3 gives 3x=y+1, k=2 requires odd=even, and k=1 gives x=3y+1.
For k=3 (x+y)=(4x1) and 16xy+1=(4x1)(12x1).
For k=1 (x+y)=(4y+1) and 16xy+1=(4y+1)(12y+1).
Verifying the gcd requirement is easy but a little messy. I expressed the two in terms of x or y then added and subtracted and divided out factors of 2 to get there.

Posted by xdog
on 20150510 14:33:17 