Metamath Proof Explorer


Theorem pellqrex

Description: There is a nontrivial solution of a Pell equation in the first quadrant. (Contributed by Stefan O'Rear, 18-Sep-2014)

Ref Expression
Assertion pellqrex DxPell1QRD1<x

Proof

Step Hyp Ref Expression
1 eldifi DD
2 eldifn D¬D
3 1 anim1i DDDD
4 fveq2 a=Da=D
5 4 eleq1d a=DaD
6 df-squarenn =a|a
7 5 6 elrab2 DDD
8 3 7 sylibr DDD
9 2 8 mtand D¬D
10 pellex D¬Dcdc2Dd2=1
11 1 9 10 syl2anc Dcdc2Dd2=1
12 simpll Dcdc2Dd2=1D
13 nnnn0 cc0
14 13 adantr cdc0
15 14 ad2antlr Dcdc2Dd2=1c0
16 nnnn0 dd0
17 16 adantl cdd0
18 17 ad2antlr Dcdc2Dd2=1d0
19 simpr Dcdc2Dd2=1c2Dd2=1
20 pellqrexplicit Dc0d0c2Dd2=1c+DdPell1QRD
21 12 15 18 19 20 syl31anc Dcdc2Dd2=1c+DdPell1QRD
22 1re 1
23 22 a1i Dcd1
24 22 22 readdcli 1+1
25 24 a1i Dcd1+1
26 nnre cc
27 26 ad2antrl Dcdc
28 1 adantr DcdD
29 28 nnrpd DcdD+
30 29 rpsqrtcld DcdD+
31 30 rpred DcdD
32 nnre dd
33 32 ad2antll Dcdd
34 31 33 remulcld DcdDd
35 27 34 readdcld Dcdc+Dd
36 22 ltp1i 1<1+1
37 36 a1i Dcd1<1+1
38 nnge1 c1c
39 38 ad2antrl Dcd1c
40 1t1e1 11=1
41 nnge1 D1D
42 sq1 12=1
43 42 a1i D12=1
44 nncn DD
45 44 sqsqrtd DD2=D
46 41 43 45 3brtr4d D12D2
47 22 a1i D1
48 nnrp DD+
49 48 rpsqrtcld DD+
50 49 rpred DD
51 0le1 01
52 51 a1i D01
53 49 rpge0d D0D
54 47 50 52 53 le2sqd D1D12D2
55 46 54 mpbird D1D
56 28 55 syl Dcd1D
57 nnge1 d1d
58 57 ad2antll Dcd1d
59 23 51 jctir Dcd101
60 lemul12a 101D101d1D1d11Dd
61 59 31 59 33 60 syl22anc Dcd1D1d11Dd
62 56 58 61 mp2and Dcd11Dd
63 40 62 eqbrtrrid Dcd1Dd
64 23 23 27 34 39 63 le2addd Dcd1+1c+Dd
65 23 25 35 37 64 ltletrd Dcd1<c+Dd
66 65 adantr Dcdc2Dd2=11<c+Dd
67 breq2 x=c+Dd1<x1<c+Dd
68 67 rspcev c+DdPell1QRD1<c+DdxPell1QRD1<x
69 21 66 68 syl2anc Dcdc2Dd2=1xPell1QRD1<x
70 69 ex Dcdc2Dd2=1xPell1QRD1<x
71 70 rexlimdvva Dcdc2Dd2=1xPell1QRD1<x
72 11 71 mpd DxPell1QRD1<x