Metamath Proof Explorer


Theorem pell1234qrreccl

Description: General solutions of the Pell equation are closed under reciprocals. (Contributed by Stefan O'Rear, 18-Sep-2014)

Ref Expression
Assertion pell1234qrreccl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ Pell1234QR ⁡ D

Proof

Step Hyp Ref Expression
1 elpell1234qr ⊢ D ∈ ℕ ∖ ◻ ℕ → A ∈ Pell1234QR ⁡ D ↔ A ∈ ℝ ∧ ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
2 1 biimpa ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → A ∈ ℝ ∧ ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
3 pell1234qrre ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → A ∈ ℝ
4 pell1234qrne0 ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → A ≠ 0
5 3 4 rereccld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ ℝ
6 5 ad2antrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℝ
7 simplrl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a ∈ ℤ
8 simplrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → b ∈ ℤ
9 8 znegcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → − b ∈ ℤ
10 5 recnd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ ℂ
11 10 ad2antrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℂ
12 zcn ⊢ a ∈ ℤ → a ∈ ℂ
13 12 adantr ⊢ a ∈ ℤ ∧ b ∈ ℤ → a ∈ ℂ
14 13 ad2antlr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a ∈ ℂ
15 eldifi ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℕ
16 15 nncnd ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℂ
17 16 ad3antrrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℂ
18 17 sqrtcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℂ
19 8 zcnd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → b ∈ ℂ
20 19 negcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → − b ∈ ℂ
21 18 20 mulcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ − b ∈ ℂ
22 14 21 addcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a + D ⁢ − b ∈ ℂ
23 3 recnd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → A ∈ ℂ
24 23 ad2antrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A ∈ ℂ
25 4 ad2antrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A ≠ 0
26 18 19 sqmuld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b 2 = D 2 ⁢ b 2
27 17 sqsqrtd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D 2 = D
28 27 oveq1d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D 2 ⁢ b 2 = D ⁢ b 2
29 26 28 eqtr2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b 2 = D ⁢ b 2
30 29 oveq2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a 2 − D ⁢ b 2 = a 2 − D ⁢ b 2
31 simprr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a 2 − D ⁢ b 2 = 1
32 18 19 mulcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b ∈ ℂ
33 subsq ⊢ a ∈ ℂ ∧ D ⁢ b ∈ ℂ → a 2 − D ⁢ b 2 = a + D ⁢ b ⁢ a − D ⁢ b
34 14 32 33 syl2anc ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a 2 − D ⁢ b 2 = a + D ⁢ b ⁢ a − D ⁢ b
35 30 31 34 3eqtr3d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 = a + D ⁢ b ⁢ a − D ⁢ b
36 24 25 recidd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A ⁢ 1 A = 1
37 simprl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A = a + D ⁢ b
38 18 19 mulneg2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ − b = − D ⁢ b
39 38 oveq2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a + D ⁢ − b = a + − D ⁢ b
40 14 32 negsubd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a + − D ⁢ b = a − D ⁢ b
41 39 40 eqtrd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a + D ⁢ − b = a − D ⁢ b
42 37 41 oveq12d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A ⁢ a + D ⁢ − b = a + D ⁢ b ⁢ a − D ⁢ b
43 35 36 42 3eqtr4d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A ⁢ 1 A = A ⁢ a + D ⁢ − b
44 11 22 24 25 43 mulcanad ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A = a + D ⁢ − b
45 sqneg ⊢ b ∈ ℂ → − b 2 = b 2
46 19 45 syl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → − b 2 = b 2
47 46 oveq2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ − b 2 = D ⁢ b 2
48 47 oveq2d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a 2 − D ⁢ − b 2 = a 2 − D ⁢ b 2
49 48 31 eqtrd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → a 2 − D ⁢ − b 2 = 1
50 oveq1 ⊢ c = a → c + D ⁢ d = a + D ⁢ d
51 50 eqeq2d ⊢ c = a → 1 A = c + D ⁢ d ↔ 1 A = a + D ⁢ d
52 oveq1 ⊢ c = a → c 2 = a 2
53 52 oveq1d ⊢ c = a → c 2 − D ⁢ d 2 = a 2 − D ⁢ d 2
54 53 eqeq1d ⊢ c = a → c 2 − D ⁢ d 2 = 1 ↔ a 2 − D ⁢ d 2 = 1
55 51 54 anbi12d ⊢ c = a → 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1 ↔ 1 A = a + D ⁢ d ∧ a 2 − D ⁢ d 2 = 1
56 oveq2 ⊢ d = − b → D ⁢ d = D ⁢ − b
57 56 oveq2d ⊢ d = − b → a + D ⁢ d = a + D ⁢ − b
58 57 eqeq2d ⊢ d = − b → 1 A = a + D ⁢ d ↔ 1 A = a + D ⁢ − b
59 oveq1 ⊢ d = − b → d 2 = − b 2
60 59 oveq2d ⊢ d = − b → D ⁢ d 2 = D ⁢ − b 2
61 60 oveq2d ⊢ d = − b → a 2 − D ⁢ d 2 = a 2 − D ⁢ − b 2
62 61 eqeq1d ⊢ d = − b → a 2 − D ⁢ d 2 = 1 ↔ a 2 − D ⁢ − b 2 = 1
63 58 62 anbi12d ⊢ d = − b → 1 A = a + D ⁢ d ∧ a 2 − D ⁢ d 2 = 1 ↔ 1 A = a + D ⁢ − b ∧ a 2 − D ⁢ − b 2 = 1
64 55 63 rspc2ev ⊢ a ∈ ℤ ∧ − b ∈ ℤ ∧ 1 A = a + D ⁢ − b ∧ a 2 − D ⁢ − b 2 = 1 → ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
65 7 9 44 49 64 syl112anc ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
66 6 65 jca ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
67 66 ex ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D ∧ a ∈ ℤ ∧ b ∈ ℤ → A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
68 67 rexlimdvva ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
69 68 adantld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → A ∈ ℝ ∧ ∃ a ∈ ℤ ∃ b ∈ ℤ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
70 2 69 mpd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
71 elpell1234qr ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 A ∈ Pell1234QR ⁡ D ↔ 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
72 71 adantr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ Pell1234QR ⁡ D ↔ 1 A ∈ ℝ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ 1 A = c + D ⁢ d ∧ c 2 − D ⁢ d 2 = 1
73 70 72 mpbird ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1234QR ⁡ D → 1 A ∈ Pell1234QR ⁡ D