*Problem*

where a and b are given real numbers.What conditions must hold on a and b for the solutions to be positive and distinct?

*Solution*

As the first step, let's express z in terms of a and b. To this end, rewrite the first equation as

and square

From the third equation we then have

and the second equation leads to b^2 = a^2 - 2az. If a=0 then also b=0 and the second equation gives x=y=z=0. If a\ne 0, we get

Next, eliminate z from the first and the third equations:

x and y

It follows that x and y are the two solutions to the quadratic equation:

Thus,

This simplifies to

First of allx and y are real and different iff (3a^2 -b^2)(3b^2 -a^2)\gt 0.

Assuming a\gt 0, they are positive iff

which is equivalent to 4(a^2 -b^2)^2\gt 0 and is always satisfied. This also shows that for {a\lt 0$} there are no solutions.

So we focus on (3a^2 -b^2)(3b^2 -a^2)\gt 0.

The factors need to be either both positive or both negative. Assuming the latter we arrive at a contradiction:

The only possibility is for the two factors to be positive:

The conclusion is that the system has distinct positive solutions iff a\gt 0 and 9a^2 \gt 3b^2 \gt a^2.