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Bivariate Quadratic Puzzle (Posted on 2015-06-14) Difficulty: 3 of 5
Find all pairs (x, y) of integers that satisfy this equation:

1 + x2y = x2 + 2xy + 2x + y

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Analytical Solution | Comment 1 of 2
Solving for y gives 
y = 1 + 4x/(x^2 - 2x -1)

X is relatively prime to x^2 - 2x -1, so either 
(a) x = 0 or 
(b) 4 is divided by x^2 - 2x -1

In the 2nd case, x^2 - 2x - 1 = p where p is in (1, 2, 4, -1, -2, -4)

Solving that quadratic equation gives
x = 1 +/- sqrt(p+2)

if p = -1, then x is 0 or 2
if p = -2 then x is 1
if p = 2  then x is -1 or 3

So, the only solutions are:
(0,1), (2,-7),  (1,-1), (-1,-1) and (3,7) 

  Posted by Steve Herman on 2015-06-14 11:24:56
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