reserve a,b,i,j,k,l,m,n for Nat;

theorem
  for a,b be Integer holds
    a*b*(a-b) divides (a-b)*(a+b)|^n -(a|^(n+1)-b|^(n+1))
  proof
    let a,b be Integer;
    A0: len ((a,b)In_Power (n+1-1)) = n+1 & len ((a,b)Subnomial (n+1-1)) = n+1;
    reconsider R1 = ((a,b)In_Power n) as Element of (n+1)-tuples_on REAL
      by A0,FINSEQ_2:92;
    reconsider R2 = ((a,b)Subnomial n) as Element of (n+1)-tuples_on REAL
      by A0,FINSEQ_2:92;
    A1: Sum (((a,b)In_Power n) - ((a,b) Subnomial n)) = Sum (R1-R2)
    .= Sum((a,b)In_Power n) - Sum ((a,b) Subnomial n) by RVSUM_1:90;
    reconsider f = (((a,b)In_Power n) - ((a,b) Subnomial n)) as
      INT-valued FinSequence;
    for i be Nat st i in dom f holds
      (a*b) divides (((a,b)In_Power n) - ((a,b) Subnomial n)).i
    proof
      let i be Nat such that
      B1: i in dom f;
      a*b divides (((a,b)In_Power n).i - ((a,b) Subnomial n).i) by DIS;
      hence thesis by B1,VALUED_1:13;
    end; then
    A2: (a*b) divides Sum (((a,b)In_Power n) - ((a,b) Subnomial n)) by INT436;
    (a-b)*(a+b)|^n -(a|^(n+1)-b|^(n+1)) =
      (a-b)*(a+b)|^n - (a-b)*Sum((a,b) Subnomial n) by SumS
    .= (a-b)*Sum((a,b)In_Power n) - (a-b)*Sum((a,b) Subnomial n) by NEWTON:30
    .= (a-b)*(Sum((a,b)In_Power n) - Sum((a,b) Subnomial n));
    hence thesis by A1,A2,NEWTON02:2;
  end;
