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   A Bivariate Divisibility Puzzle
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   Author  Topic: A Bivariate Divisibility Puzzle  (Read 676 times)
K Sengupta
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A Bivariate Divisibility Puzzle  
« on: May 30th, 2006, 8:22pm »
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Determine all possible pairs (x,y) of positive integers in order that (x^2)/( 8*x*y^2 – 8*y^3 +1) is a positive integer.
« Last Edit: May 30th, 2006, 8:23pm by K Sengupta » IP Logged
K Sengupta
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Re: A Bivariate Divisibility Puzzle  
« Reply #1 on: Jun 3rd, 2006, 8:18am »
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Let us substitute h = x^2/(8*x*y^2 – 8*y^3 +1). Then we have a quadratic equation for m, namely x^2 – 8*h*y^2*x + (8*y^3 - 1)h = 0. This has solutions 4h*y^2 +/- N, where N is the positive square root of 16*h^2*y^4 - 8* h*y^3 + h.  
 
Since y >= 1, h >= 1, N is certainly real. But the sum and product of the roots are both positive, so both roots must be positive. The sum is an integer, so if one root is a positive integer, then so is the other.  
 
The larger root 4h*y^2 + N is greater than 4h*y^2, so the smaller root < h(8*y^3 - 1)/(4*h*y^2) < 2y.So, we can't have y < x < 2y, for otherwise: 8(x-y)y^2 + 1 > (2y)^2 > x^2.
So x < 2y implies x< =y.  
 
So for the smaller root we cannot have x-y > 0. But x-y must be non-negative (since h is positive), so x-y = 0 for the smaller root. Hence 4*h*y^2 - N = y.  
 
Now N^2 = (4*h*y^2 - y)^2 = 16*h^2*y^4 -8* h*y^3 + h, so h = y^2. Substituting y = k we obtain the solutions: (x,y) = (k, k) and (8*k^4 - k, k).
 
We have shown that any solution must be of one of the two forms given, but it is trivial to check that they are all indeed solutions.
« Last Edit: Jun 8th, 2006, 7:45am by K Sengupta » IP Logged
Eigenray
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Re: A Bivariate Divisibility Puzzle  
« Reply #2 on: Jun 3rd, 2006, 11:10pm »
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Interesting.  Ignoring the "positive" part, the solutions (x,y) are in bijection with divisors d | (8y3-1)2 that are 1 mod 8y2, via d = 8y2(x-y)+1.
 
Since
(8y2x - 8y3 + 1)(8y2x + 8y3 - 1) = (8y2x)2 - (8y3-1)2,
and d | x2, we have d | (8y3-1)2, and clearly d = 1 mod 8y2.
 
Conversely, if d | (8y3-1)2, then d | (8y2x)2; since d is relatively prime to 2y, we have d | x2, and if also d = 1 mod 8y2, we can solve for x = y + (d-1)/(8y2).
 
The three "trivial" solutions are d = 1, (1-8y3), and (8y3-1)2, we correspond to x = y, 0, and y(8y3-1), respectively.  Are there any others?  That is, if z is even, and d | (z3-1)2, and d = 1 mod 2z2, must d be either 1, (1-z3), or (z3-1)2?  Equivalently, when allowing x,y to be arbitrary integers, are any new solutions added except x=0 or y=0?
 
on Jun 3rd, 2006, 8:18am, K Sengupta wrote:
But note that if x-y > 0, then since h > 0, we must have the denominator 8*(x-y)*(y^2 )+ 1 smaller than the numerator and hence x>y.

I don't understand "hence x>y" here.  Perhaps a better phrasing is: we can't have y < x < 2y, for otherwise
8(x-y)y2 + 1 > (2y)2 > x2.
So x < 2y implies x<y.
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K Sengupta
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Re: A Bivariate Divisibility Puzzle  
« Reply #3 on: Jun 8th, 2006, 7:58am »
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On Jun 3rd, 2006, 11:10pm, Eigenray wrote:
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I don't understand "hence x>y" here.

Incorporated the necessary amendments in my previous post in conformity with the abovementioned suggestion.
 Also, corrected some typographical anomalies is the said post. Consequently, the said post is now fit for public display.
Thanks, Eigenray.
 
On Jun 3rd, 2006, 11:10pm, Eigenray wrote:
Quote:
Are there any others?  That is, if z is even, and d | (z3-1)2, and d = 1 mod 2z2, must d be either 1, (1-z3), or (z3-1)2?  Equivalently, when allowing x,y to be arbitrary integers, are any new solutions added except x=0 or y=0?

 
Despite my best efforts I have not been able to determine any further solutions, and accordingly, I am still looking for them.
« Last Edit: Jun 8th, 2006, 8:01am by K Sengupta » IP Logged
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