
theorem
  {n where n is positive Integer:
  5 divides 4*(n|^2) + 1 & 13 divides 4*(n|^2) + 1} is infinite
  proof
    set S={n where n is positive Integer:
    5 divides 4*n|^2 + 1 & 13 divides 4*n|^2 + 1};
    deffunc F(Nat) = 65*$1+56;
    consider f being ManySortedSet of NAT such that
    f: for n being Element of NAT holds f.n=F(n) from PBOOLE:sch 5;
    set R = rng f;
    d: dom f = NAT by PARTFUN1:def 2;
    rs: R c= S
    proof
      let a be object;
      assume a in R;
      then consider k being object such that
      k: k in dom f & a=f.k by FUNCT_1:def 3;
      reconsider k as Element of NAT by d,k;
      a: a=65*k+56 by f,k;
      then reconsider n=a as positive Integer;
      c: 4*(n|^(1+1)) + 1 = 4*((n|^(1))*n)+1 by NEWTON:6
      .= 4*(n*n) + 1 by NEWTON:5;
      65=5*13; then
   p: 13 divides 65-0 by INT_1:def 3;
      65*k+56-1 = 5*(13*k+11);
      then 5 divides 65*k+56-1 by INT_1:def 3;
      then n,1 are_congruent_mod 5 by a,INT_1:def 4;
      then n*n,1*1 are_congruent_mod 5 & 4,4 are_congruent_mod 5
      by INT_1:18,11;
      then 4*(n*n),1*4 are_congruent_mod 5 & 1,1 are_congruent_mod 5
      by INT_1:18,11;
      then 4*(n*n)+1,4+1 are_congruent_mod 5 & 5,0 are_congruent_mod 5
      by INT_1:16,12;
      then 4*(n*n)+1,0 are_congruent_mod 5 by INT_1:15;
      then 4*n|^2 + 1,0 are_congruent_mod 5 by c;
      then d1: 5 divides (4*n|^2 + 1) - 0 by INT_1:def 4;
      65*k+56-4 = 13*(5*k+4);
      then 13 divides 65*k+56-4 by INT_1:def 3;
      then n,4 are_congruent_mod 13 by a,INT_1:def 4;
      then n*n,4*4 are_congruent_mod 13 & 4,4 are_congruent_mod 13
      by INT_1:18,11;
      then 4*(n*n),4*4*4 are_congruent_mod 13 & 1,1 are_congruent_mod 13
      by INT_1:18,11;
      then 4*(n*n)+1,64+1 are_congruent_mod 13 & 65,0 are_congruent_mod 13
      by INT_1:16,def 4,p;
      then 4*(n*n)+1,0 are_congruent_mod 13 by INT_1:15;
      then 4*n|^2 + 1,0 are_congruent_mod 13 by c;
      then 13 divides (4*n|^2 + 1) - 0 by INT_1:def 4;
      hence a in S by d1;
    end;
    for m being Nat ex n being Nat st n >= m & n in R
    proof
      let m be Nat;
      take n=F(m);
      65*m >= 1*m by XREAL_1:66;
      hence n >= m by XREAL_1:38;
aa:   m in NAT by ORDINAL1:def 12;
      then m in dom f by d;
      then f.m in rng f by FUNCT_1:3;
      hence n in R by f,aa;
    end;
    then R is infinite by PYTHTRIP:9;
    hence thesis by rs;
  end;
