A typical question from 1989 leningrad mathematical olympiad

Prove that we cannot define an operation * on the set of integers Z satisfy all of the three properties below simultaneously:

For any A∈Z,B∈Z,C∈Z: 1.AB=-(BA) 2.(A*B)C=A(BC) (Associative Law) 3.For every A∈Z,there exist B∈Z,C∈Z such that A=BC

I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:

1.for any X∈Z,we have XX=-(XX).So we have X*X=0

2.for any X∈Z,we have X*0=-(0*X)=X*(XX)=(XX)*X=0*X.So we have X*0=0*X=0

3.Now For any X∈Z(X not 0).We define the orbit of X–Ox to be the set Ox={S|∃Y∈Z such that XY=S},and the stabilizer of X-Fx to be the set Fx={T|TX=X*T=0}.

My goal is to prove that actually Ox=Fx.And therefore since X∈Fx,so X∈Ox,and we reach a contradiction since X∉Ox(otherwise if ∃Y∈Z such that XY=X,then (XY)Y=X(Y*Y)=X*0=X*Y=X=0)

It is easy to see that Ox⊆Fx,since for any S∈Ox,we have XS=X(XY)=(XX)*Y=0*Y=0,so S∈Fx

However,for the other part,I cannot deduce out,which I need help.