Metamath Proof Explorer


Theorem pell1qr1

Description: 1 is a Pell solution and in the first quadrant as one. (Contributed by Stefan O'Rear, 17-Sep-2014)

Ref Expression
Assertion pell1qr1 ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 ∈ Pell1QR ⁡ D

Proof

Step Hyp Ref Expression
1 1red ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 ∈ ℝ
2 1nn0 ⊢ 1 ∈ ℕ 0
3 2 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 ∈ ℕ 0
4 0nn0 ⊢ 0 ∈ ℕ 0
5 4 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ → 0 ∈ ℕ 0
6 eldifi ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℕ
7 6 nncnd ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℂ
8 7 sqrtcld ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℂ
9 8 mul01d ⊢ D ∈ ℕ ∖ ◻ ℕ → D ⋅ 0 = 0
10 9 oveq2d ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 + D ⋅ 0 = 1 + 0
11 1p0e1 ⊢ 1 + 0 = 1
12 10 11 eqtr2di ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 = 1 + D ⋅ 0
13 sq1 ⊢ 1 2 = 1
14 13 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 2 = 1
15 sq0 ⊢ 0 2 = 0
16 15 oveq2i ⊢ D ⁢ 0 2 = D ⋅ 0
17 7 mul01d ⊢ D ∈ ℕ ∖ ◻ ℕ → D ⋅ 0 = 0
18 16 17 eqtrid ⊢ D ∈ ℕ ∖ ◻ ℕ → D ⁢ 0 2 = 0
19 14 18 oveq12d ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 2 − D ⁢ 0 2 = 1 − 0
20 1m0e1 ⊢ 1 − 0 = 1
21 19 20 eqtrdi ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 2 − D ⁢ 0 2 = 1
22 oveq1 ⊢ a = 1 → a + D ⁢ b = 1 + D ⁢ b
23 22 eqeq2d ⊢ a = 1 → 1 = a + D ⁢ b ↔ 1 = 1 + D ⁢ b
24 oveq1 ⊢ a = 1 → a 2 = 1 2
25 24 oveq1d ⊢ a = 1 → a 2 − D ⁢ b 2 = 1 2 − D ⁢ b 2
26 25 eqeq1d ⊢ a = 1 → a 2 − D ⁢ b 2 = 1 ↔ 1 2 − D ⁢ b 2 = 1
27 23 26 anbi12d ⊢ a = 1 → 1 = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 ↔ 1 = 1 + D ⁢ b ∧ 1 2 − D ⁢ b 2 = 1
28 oveq2 ⊢ b = 0 → D ⁢ b = D ⋅ 0
29 28 oveq2d ⊢ b = 0 → 1 + D ⁢ b = 1 + D ⋅ 0
30 29 eqeq2d ⊢ b = 0 → 1 = 1 + D ⁢ b ↔ 1 = 1 + D ⋅ 0
31 oveq1 ⊢ b = 0 → b 2 = 0 2
32 31 oveq2d ⊢ b = 0 → D ⁢ b 2 = D ⁢ 0 2
33 32 oveq2d ⊢ b = 0 → 1 2 − D ⁢ b 2 = 1 2 − D ⁢ 0 2
34 33 eqeq1d ⊢ b = 0 → 1 2 − D ⁢ b 2 = 1 ↔ 1 2 − D ⁢ 0 2 = 1
35 30 34 anbi12d ⊢ b = 0 → 1 = 1 + D ⁢ b ∧ 1 2 − D ⁢ b 2 = 1 ↔ 1 = 1 + D ⋅ 0 ∧ 1 2 − D ⁢ 0 2 = 1
36 27 35 rspc2ev ⊢ 1 ∈ ℕ 0 ∧ 0 ∈ ℕ 0 ∧ 1 = 1 + D ⋅ 0 ∧ 1 2 − D ⁢ 0 2 = 1 → ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 1 = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
37 3 5 12 21 36 syl112anc ⊢ D ∈ ℕ ∖ ◻ ℕ → ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 1 = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
38 elpell1qr ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 ∈ Pell1QR ⁡ D ↔ 1 ∈ ℝ ∧ ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 1 = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
39 1 37 38 mpbir2and ⊢ D ∈ ℕ ∖ ◻ ℕ → 1 ∈ Pell1QR ⁡ D