It is necessary to assume that x1 and x0 are known. This makes the fact that x2=x1-x0 given.
We write
xn+1=xn-xn-1
xn+2=xn+1-xn=xn-xn-1-xn=-xn-1
xn+3=xn+2-xn+1=-xn
xn+4=xn+3-xn+2=-xn+xn-1
xn+5=xn+4-xn+3=xn-1
xn+6=xn+5-xn+4=xn
xn+7=xn+6-xn+5=xn-xn-1=xn+1 we are home!
Kyle C.
05/03/16