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 Th70:
  2|^m - 2|^n = 2 implies m = 2 & n = 1
  proof
    assume
A1: 2|^m - 2|^n = 2;
    now
      assume n >= m;
      then 2|^m - 2|^n <= 2|^n - 2|^n by XREAL_1:9,PREPOWER:93;
      hence contradiction by A1;
    end;
    then consider k such that
A2: m = n+k by NAT_1:10;
    per cases;
    suppose 1 <= n;
      then reconsider n1 = n-1 as Element of NAT by INT_1:3;
A3:   2|^(n+k) = 2|^n * 2|^k by NEWTON:8;
      2|^n = 2|^(n1+1)
      .= 2|^n1*2 by NEWTON:6;
      then 2|^n1*2*(2|^k-1) = 2*1 by A1,A2,A3;
      then 2|^n1*(2|^k-1) = 1;
      then 2|^n1 = 1 & 2|^k-1 = 1 by NAT_1:15;
      then 2|^n1 = 2|^0 & 2|^k = 2|^1 by NEWTON:4;
      then n1 = 0 & k = 1 by PEPIN:30;
      hence thesis by A2;
    end;
    suppose n < 1;
      then
A4:   n = 0 by NAT_1:14;
      then m = 0 by A1;
      hence thesis by A1,A4;
    end;
  end;
