Re: Diophantine equation
- From: Timothy Murphy <tim@xxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 24 Feb 2006 13:46:10 +0000
Gaurav1146 wrote:
I was trying to solve the following question and came across an
interesting result though I could not prove it.
Find all possible positive integers x and y such that x^2+y^2+1 is a
multiple of (x*y).
This can be written as x^2+y^2+1 = N*x*y where x,y and N are positive
integers.
I wrote a java code to generate all such pairs. What I observed was
that solutions existed for N=3 only, though I couldnt prove it. The
fact that N should be a multiple of 3 can be easily proved.
I think N must be odd.
For if N is even just one of x,y must be even,
and then x^2 + y^2 + 1 = 2 mod 4, while Nxy = 0 mod 4.
I would try re-writing the equation (for given N) as
(2x - Ny)^2 - (N^2 - 4)y^2 = -4,
say
u^2 - dv^2 = -4,
where d = N^2 - 4.
There is obviously no solution if N = 0,1,2.
So d > 0.
Evidently -1 must be a quadratic residue mod d.
I would hazard a guess that this is sufficient for there to be a solution.
If N = 7 then d = 45 = 5 x 9,
so -1 is a quadratic residue mod d.
Is there a solution in this case?
The answer is left to the reader ...
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
.
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