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Divisibility to Square II (Posted on 2016-08-07) Difficulty: 4 of 5
This is a followup to Divisibility to square.

Each of x and y is a positive integer such that x^2 + y^2 + x is divisible by xy.

1: Prove that there is an infinite number of (x,y) which make the quotient equal to 3.

2: Prove or disprove 3 is the only integer quotient possible.

No Solution Yet Submitted by Brian Smith    
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Some Thoughts Possible solution Comment 2 of 2 |

Assuming the quotient, q, is positive (if the quotient is negative, x can never be positive for the equality x^2 + nxy + y^2 + x = 0 to hold):

Let x^2 - xy + y^2 + x = 0, an ellipse with integer solutions (x,y}={-1,-1},{0,0},{-1,0}  
Let x^2 - 2xy + y^2 + x = 0,  a parabola with integer solutions {x,y} ={-u^2, (u^2-u)} 
Let x^2-3xy+y^2+x= 0, a hyperbola with integer solutions {x,y} = {0,0}, {-1,0}, {1,1}, {1,2}, {4,2}; plainly there are infinitely many such solutions of the form:  
Xn+1 = P Xn + Q Yn + K 
Yn+1 = R Xn + S Yn + L 
with characteristics
P = -3 
Q = 1 
K = 1 
R = -1 
S = 0 
L = 1
or 
P = 0
Q = -1
K = 1
R = 1
S = -3
L = 2

If the quotient is greater than 3, then the characteristics are of the form:

P = (q^2-1) 
Q = -q 
K = -1 
R = q 
S = -1 
L = 0 
or
P = -1
Q = q
K = -1
R = -q
S = (q^2-1)
L = -q

Since the terms in P and Q, and in R and S, are of opposite sign, either x or y is never positive, contrary to the stipulations of the problem. So 3 is the only integer quotient possible, as was to be proved.

 

Edited on August 8, 2016, 7:12 am
  Posted by broll on 2016-08-08 00:30:54

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