reserve a,b,c,d,x,j,k,l,m,n,o,xi,xj for Nat,
  p,q,t,z,u,v for Integer,
  a1,b1,c1,d1 for Complex;

theorem
  for p be prime Nat, a,b be Integer st |.a.| <> |.b.| holds
    p |-count (a|^2 - b|^2) = (p |-count (a-b)) + (p|-count (a+b))
  proof
    let p be prime Nat, a,b be Integer such that
A1: |.a.| <> |.b.|;
    a - b <> 0 & a + b <> 0 by A1,ABS1; then
    reconsider t = |.a - b.|, u = |.a + b.| as non zero Nat;
    p |-count (a|^2 - b|^2) = p|-count |.(a+b)*(a-b).| by NEWTON01:1
    .= p |-count (t*u) by COMPLEX1:65
    .= (p |-count |.a-b.|)+ (p|-count |.a+b.|) by NAT_3:28;
    hence thesis;
  end;
