let D, m be positive Integer; ( D = (m ^2) + 1 implies { [x,y] where x, y is positive Integer : (x ^2) - (D * (y ^2)) = 1 } is infinite )
assume A1:
D = (m ^2) + 1
; { [x,y] where x, y is positive Integer : (x ^2) - (D * (y ^2)) = 1 } is infinite
set f = exampleSierpinski150 (m,D);
defpred S1[ Complex, Complex] means ($1 ^2) - (D * ($2 ^2)) = 1;
set A = { [x,y] where x, y is positive Integer : S1[x,y] } ;
A2:
S1[(2 * (m ^2)) + 1,2 * m]
by A1;
then
[((2 * (m ^2)) + 1),(2 * m)] in { [x,y] where x, y is positive Integer : S1[x,y] }
;
then reconsider A = { [x,y] where x, y is positive Integer : S1[x,y] } as non empty set ;
deffunc H5( Real, Real) -> set = ($1 ^2) + (D * ($2 ^2));
deffunc H6( Real, Real) -> set = (2 * $1) * $2;
A3:
dom (exampleSierpinski150 (m,D)) = NAT
by PARTFUN1:def 2;
defpred S2[ Nat] means (exampleSierpinski150 (m,D)) . $1 in A;
(exampleSierpinski150 (m,D)) . 0 = [((2 * (m ^2)) + 1),(2 * m)]
by Def19;
then A4:
S2[ 0 ]
by A2;
A5:
for a being Nat st S2[a] holds
S2[a + 1]
proof
let a be
Nat;
( S2[a] implies S2[a + 1] )
assume
S2[
a]
;
S2[a + 1]
then consider x,
y being
positive Integer such that A6:
(
(exampleSierpinski150 (m,D)) . a = [x,y] &
S1[
x,
y] )
;
A7:
(exampleSierpinski150 (m,D)) . (a + 1) = [H5(((exampleSierpinski150 (m,D)) . a) `1 ,((exampleSierpinski150 (m,D)) . a) `2 ),H6(((exampleSierpinski150 (m,D)) . a) `1 ,((exampleSierpinski150 (m,D)) . a) `2 )]
by Def19;
(((x ^2) + (D * (y ^2))) ^2) - (D * (((2 * x) * y) ^2)) = ((x ^2) - (D * (y ^2))) ^2
;
hence
S2[
a + 1]
by A6, A7;
verum
end;
A8:
for a being Nat holds S2[a]
from NAT_1:sch 2(A4, A5);
rng (exampleSierpinski150 (m,D)) c= A
hence
{ [x,y] where x, y is positive Integer : (x ^2) - (D * (y ^2)) = 1 } is infinite
by A3, CARD_1:59; verum