Hi,

I have this system of equations

$\displaystyle x^2>x^2-4xy+2xz+2y^2-z^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

$\displaystyle x^2-4xy+2xz+2y^2-z^2>x^2-8xy+4xz+4y^2-2z^2\ \ \ \ \ \ \ \ \ (2)$

$\displaystyle x^2-8xy+4xz+4y^2-2z^2>x^2+2xy-2xz-y^2+z^2\ \ \ \ \ \ \ \ \ \ \ (3)$

$\displaystyle x^2+2xy-2xz-y^2+z^2>x^2-2xy+y^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

$\displaystyle x^2-2xy+y^2>x^2-6xy+2xz+3y^2-z^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

$\displaystyle x^2-6xy+2xz+3y^2-z^2>x^2+4xy-4xz-2y^2+2z^2\ \ \ \ \ \ \ \ \ \ \ (6)$

$\displaystyle x^2+4xy-4xz-2y^2+2z^2>x^2-2xz+z^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7)$

$\displaystyle x^2-2xz+z^2>x^2-4xy+2y^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)$

$\displaystyle x^2-4xy+2y^2>0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (9)$

From here on I'm only working with +ve x, y and z integers where $\displaystyle \ \ \ \ \ \ \ x\ge4>z>y>0$

For any given x i've found the maximum value of y is

$\displaystyle x\left(1-\dfrac{1}{\sqrt{2}}\right)$

and that for y from 1 to this limit, z has a lower range and higher range of limits

$\displaystyle \dfrac{3x+3-\sqrt{3(3x^2-10xy+5y^2}}{3}\ \ \ \le\ z\ \le\ \ \ x-\sqrt{x^2-4xy+2y^2}$

and

$\displaystyle x+1+\sqrt{x^2-4xy+2y^2}\ \ \ \le\ z\ \le\ \ \ \dfrac{3x+\sqrt{3(3x^2-10xy+5y^2}}{3}$

I had thought that y had a lower limit of 1

but when x gets to 31622777 and y = 1 there are no valid values of Z that satisfy the equations.

Therearevalid values for z when x = 31622777 and y = 2

So somewhere in these equations I should be able to find this lower y limit

My problem? I can't

Any help would be greatly appreciated

Pro