
theorem
  for x being Integer st x-3 divides x|^3 - 3 holds
  x=-21 or x=-9 or x=-5 or x=-3 or x=-1 or x=0 or x=1 or x=2 or x=4 or x=5
  or x=6 or x=7 or x=9 or x=11 or x=15 or x=27
  proof
    let x be Integer;
    assume x: x-3 divides x|^3 - 3;
    set t=x-3;
    e: (t+3)|^(2+1) - 3 = ((t+3)|^(1+1))*(t+3) - 3 by NEWTON:6
    .= (((t+3)|^1)*(t+3))*(t+3) - 3 by NEWTON:6
    .= (t+3)*(t+3)*(t+3) - 3 by NEWTON:5
    .= ((t*t*t)+(27*t)+(9*t*t))+24;
    t: t divides t*t*t & t divides 27*t & t divides 9*t*t by INT_1:def 3;
    then t divides (t*t*t)+(27*t) by WSIERP_1:4;
    then t divides ((t*t*t)+(27*t)+(9*t*t)) by t,WSIERP_1:4;
    then t divides 24 by x,e,INT_2:1;
    then
    t=-1 or t=1 or t=-2 or t=2 or t=-3 or t=3 or t=-4 or t=4 or t=-6 or t=6
    or t=-8 or t=8 or t=-12 or t=12 or t=-24 or t=24 by lem24;
    hence
    x=-21 or x=-9 or x=-5 or x=-3 or x=-1 or x=0 or x=1 or x=2 or x=4 or x=5
    or x=6 or x=7 or x=9 or x=11 or x=15 or x=27;
  end;
