because
  x*x=x (mod pq)
so
  x*x=kpq+x
  x(x-1)=kpq

p, q are distinct prime, so

case 1
  p|x, q|x-1
case 2
  p|x-1, q|x

for the first case, we have
pa=x, qb=x-1, so pa-qb=1
...