Question 1092331
<br>A very unusual kind of problem.... I hope you have shown it correctly.<br>
You show
{{{x/z = -z}}}
which means
{{{x = -z^2}}}  (1)<br>
You also show
{{{x/y = z}}}
which means
{{{x = yz}}}  (2)<br>
Then (1) and (2) together mean
{{{-z^2 = yz}}}
which means
{{{y = -z}}}  (3)<br>
So y and z are opposites; and you also show that z < y.  That means z is some negative number, call it -a, and y is then the positive number a.  (2) then tells us that x is the negative number -a^2.<br>
In summary, we have, where a is some positive number,
y = a
z = -a
x = -a^2<br>
These parametric values for x, y, and z satisfy all the given conditions but one -- that z/2 and z/3 are both integers.  If z/2 and z/3 are both integers, then z must be a multiple of 2*3 = 6.<br>
So our final result is an infinite set of solutions where
y is some negative integer multiple of 6;
z is the opposite of y; and
x is the product of y and z