modulus equation

Let a=|x|, b=|y|, c=|z|

Then we have

\begin{matrix} a^2-a= bc\\ b^2-b = ac \\ c^2-c = ab \end{matrix}

Remember that a,b,c>0

Then we can write the inequalities

\begin{matrix} a^2\ge bc\\ b^2\ge ac \\ c^2 \ge ab \end{matrix}

WLOG a \ge b \ge c

If any of the inequalitiies here becomes a strict inequality, then c^2 \ge ab becomes absurd.

(or equivalently if either a²>bc or b²>ac or c²>ab then multiplying you get the absurdity (abc)^2 > abc

So we must have a =b=c =0

@mavarick

u picked this q from titu's Mathematical Olympiad challenges ?

Continuing from Kaymant Sir's post:

That brings to the fore that some of the reasoning in #2 is wrong.

a=b=c = 0 is a trivial solution

As in #2, we have a^2 \ge bc; b^2 \ge ca ; c^2 \ge ab

If none of a,b,c are zero and one of the inequalities is a strict inequality, then on multiplication we get (abc)^2>a^2b^2c^2 which is absurd.

So we must have a^2=bc; b^2=ac;c^2=ab

which leads us to back a=b=c = 0

Now WLOG suppose a= 0. Then at least one of b or c is zero. So lets say b =0.

Then if c≠0, we must have c=1.

Hence (0,0,±1) and permutations are solutions

Yeah, I checked up, the question is from that book, though he wouldnt have picked it up from there as solutions are given (it is pretty much on the same lines)