
theorem for a,b be odd Integer holds a-b is square implies not a + b is square
  proof
    let a,b be odd Integer;
    assume a-b is square; then
    consider c be Nat such that
    A1: c^2 = a - b by PYTHTRIP:def 3;
    reconsider c as even Nat by A1;
    A2: |.a.| in NAT & |.b.| in NAT by INT_1:3; then
    reconsider k = max (|.a.|,|.b.|) as odd Nat by XXREAL_0:def 10;
    reconsider l = min (|.a.|,|.b.|) as odd Nat by A2, XXREAL_0:def 9;
    A3: |.a.| + |.b.| = k + l by NEWTON03:41;
    2 divides c by ABIAN:def 1; then
    2*2 divides c*c by NEWTON02:2; then
    A5: 2*2 divides c^2 by SQUARE_1:def 1;
    per cases by ABS;
    suppose
      B1: |.a + b.| = |.a.| + |.b.|; then
      B2: |.a - b.| = |.|.a.| - |.b.|.| by LABS
      .= k - l by MMD;
      not 4 divides k|^1 + l|^1 by A1,A5,B2,NEWTON03:20;
      hence thesis by B1,A3,A4I;
    end;
    suppose
      B1: |.a - b.| = |.a.| + |.b.|; then
      B2:  |.a + b.| = |.|.a.| - |.b.|.| by LmABS
      .= k - l by MMD;
      4 divides k|^1 + l|^1 by B1,A1,NEWTON03:41,A5; then
      not 4 divides k - l by NEWTON03:20;
      hence thesis by B2,A4I;
    end;
  end;
