reserve k,n,m,l,p for Nat;
reserve n0,m0 for non zero Nat;
reserve f for FinSequence;
reserve x,X,Y for set;
reserve f1,f2,f3 for FinSequence of REAL;
reserve n1,n2,m1,m2 for Nat;
reserve I,j for set;
reserve f,g for Function of I, NAT;
reserve J,K for finite Subset of I;

theorem Th33:
  sigma n0 = n0 + m & m divides n0 & n0<>m implies m = 1 & n0 is prime
proof
  assume
A1: sigma n0 = n0 + m;
  assume
A2: m divides n0;
  assume
A3: n0 <> m;
  per cases;
  suppose
    n0 = 1;
    then 1 = 1 + m by A1,Th29;
    hence thesis by A2,INT_2:3;
  end;
  suppose
A4: n0 <> 1;
    consider k be Nat such that
A5: n0 = m * k by A2,NAT_D:def 3;
A6: k <> m
    proof
      assume k = m;
      then m <> 1 by A4,A5;
      then 1 + m + n0 <= sigma n0 by A2,A3,Th31;
      then 1 + m + n0 - (m + n0) <= n0 + m - (m + n0) by A1,XREAL_1:9;
      hence contradiction;
    end;
    k = n0
    proof
A7:   k divides n0 & k <> 1 by A3,A5,NAT_D:def 3;
      assume
A8:   k <> n0;
      then m<>1 by A5;
      then 1 + m + k + n0 <= sigma n0 by A2,A3,A6,A8,A7,Th32;
      then 1 + m + k + n0 - (m + n0) <= n0 + m - (m + n0) by A1,XREAL_1:9;
      then 1 + k <= 0;
      hence contradiction;
    end;
    then n0/n0 = m by A5,XCMPLX_1:89;
    hence
A9: m = 1 by XCMPLX_1:60;
A10: for k being Nat holds not k divides n0 or k = 1 or k = n0
    proof
      let k be Nat;
      assume that
A11:  k divides n0 and
A12:  k <> 1;
      assume k <> n0;
      then 1 + k + n0 <= n0 + 1 by A1,A9,A11,A12,Th31;
      then 1 + k + n0 - (1 + n0) <= n0 + 1 - (1 + n0) by XREAL_1:9;
      then k = 0;
      hence contradiction by A11,INT_2:3;
    end;
    n0 > 1 by A4,NAT_1:25;
    hence n0 is prime by A10,INT_2:def 4;
  end;
end;
