reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r,s for Real;
reserve p,p1,p2,p3 for Prime;

theorem Th67:
  n <> 0 & 2|^m - 3|^n = 1 implies m = 2 & n = 1
  proof
    assume that
A1: n <> 0 and
A2: 2|^m - 3|^n = 1;
A3: 3|^0 = 1 by NEWTON:4;
    per cases by NAT_1:13;
    suppose m <= 2;
      then m = 0 or ... or m = 2;
      then per cases;
      suppose m = 0;
        then 2|^0 - 3|^n = 1 by A2;
        hence thesis by Lm1;
      end;
      suppose m = 1;
        then 2|^1 - 3|^n = 1 by A2;
        hence thesis by A1,A3,PEPIN:30;
      end;
      suppose
A4:     m = 2;
        then 2|^2 - 3|^n = 1 by A2;
        then 3|^n = 3|^1 by Lm3;
        hence thesis by A4,PEPIN:30;
      end;
    end;
    suppose
A5:   m >= 2+1;
      not 8 divides 3|^n+1 by Th66;
      hence thesis by A2,A5,Lm12,NEWTON:89;
    end;
  end;
