Metamath Proof Explorer


Theorem 2sqreunnltlem

Description: Lemma for 2sqreunnlt . (Contributed by AV, 4-Jun-2023) Specialization to different integers, proposed by GL. (Revised by AV, 11-Jun-2023)

Ref Expression
Assertion 2sqreunnltlem ⊢ P ∈ ℙ ∧ P mod 4 = 1 → ∃! a ∈ ℕ ∃! b ∈ ℕ a < b ∧ a 2 + b 2 = P

Proof

Step Hyp Ref Expression
1 2sqreunnlem1 ⊢ P ∈ ℙ ∧ P mod 4 = 1 → ∃! a ∈ ℕ ∃! b ∈ ℕ a ≤ b ∧ a 2 + b 2 = P
2 oveq1 ⊢ b = a → b 2 = a 2
3 2 oveq2d ⊢ b = a → a 2 + b 2 = a 2 + a 2
4 3 adantr ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + b 2 = a 2 + a 2
5 nncn ⊢ a ∈ ℕ → a ∈ ℂ
6 5 sqcld ⊢ a ∈ ℕ → a 2 ∈ ℂ
7 2times ⊢ a 2 ∈ ℂ → 2 ⁢ a 2 = a 2 + a 2
8 7 eqcomd ⊢ a 2 ∈ ℂ → a 2 + a 2 = 2 ⁢ a 2
9 6 8 syl ⊢ a ∈ ℕ → a 2 + a 2 = 2 ⁢ a 2
10 9 adantl ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ → a 2 + a 2 = 2 ⁢ a 2
11 10 ad2antrl ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + a 2 = 2 ⁢ a 2
12 4 11 eqtrd ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + b 2 = 2 ⁢ a 2
13 12 eqeq1d ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + b 2 = P ↔ 2 ⁢ a 2 = P
14 oveq1 ⊢ P = 2 ⁢ a 2 → P mod 4 = 2 ⁢ a 2 mod 4
15 14 eqeq1d ⊢ P = 2 ⁢ a 2 → P mod 4 = 1 ↔ 2 ⁢ a 2 mod 4 = 1
16 eleq1 ⊢ P = 2 ⁢ a 2 → P ∈ ℙ ↔ 2 ⁢ a 2 ∈ ℙ
17 15 16 anbi12d ⊢ P = 2 ⁢ a 2 → P mod 4 = 1 ∧ P ∈ ℙ ↔ 2 ⁢ a 2 mod 4 = 1 ∧ 2 ⁢ a 2 ∈ ℙ
18 nnz ⊢ a ∈ ℕ → a ∈ ℤ
19 2nn0 ⊢ 2 ∈ ℕ 0
20 zexpcl ⊢ a ∈ ℤ ∧ 2 ∈ ℕ 0 → a 2 ∈ ℤ
21 18 19 20 sylancl ⊢ a ∈ ℕ → a 2 ∈ ℤ
22 2mulprm ⊢ a 2 ∈ ℤ → 2 ⁢ a 2 ∈ ℙ ↔ a 2 = 1
23 21 22 syl ⊢ a ∈ ℕ → 2 ⁢ a 2 ∈ ℙ ↔ a 2 = 1
24 oveq2 ⊢ a 2 = 1 → 2 ⁢ a 2 = 2 ⋅ 1
25 2t1e2 ⊢ 2 ⋅ 1 = 2
26 24 25 eqtrdi ⊢ a 2 = 1 → 2 ⁢ a 2 = 2
27 26 oveq1d ⊢ a 2 = 1 → 2 ⁢ a 2 mod 4 = 2 mod 4
28 2re ⊢ 2 ∈ ℝ
29 4nn ⊢ 4 ∈ ℕ
30 nnrp ⊢ 4 ∈ ℕ → 4 ∈ ℝ +
31 29 30 ax-mp ⊢ 4 ∈ ℝ +
32 0le2 ⊢ 0 ≤ 2
33 2lt4 ⊢ 2 < 4
34 modid ⊢ 2 ∈ ℝ ∧ 4 ∈ ℝ + ∧ 0 ≤ 2 ∧ 2 < 4 → 2 mod 4 = 2
35 28 31 32 33 34 mp4an ⊢ 2 mod 4 = 2
36 27 35 eqtrdi ⊢ a 2 = 1 → 2 ⁢ a 2 mod 4 = 2
37 36 eqeq1d ⊢ a 2 = 1 → 2 ⁢ a 2 mod 4 = 1 ↔ 2 = 1
38 1ne2 ⊢ 1 ≠ 2
39 eqcom ⊢ 2 = 1 ↔ 1 = 2
40 eqneqall ⊢ 1 = 2 → 1 ≠ 2 → a ≤ b → b ≠ a
41 40 com12 ⊢ 1 ≠ 2 → 1 = 2 → a ≤ b → b ≠ a
42 39 41 biimtrid ⊢ 1 ≠ 2 → 2 = 1 → a ≤ b → b ≠ a
43 38 42 ax-mp ⊢ 2 = 1 → a ≤ b → b ≠ a
44 37 43 biimtrdi ⊢ a 2 = 1 → 2 ⁢ a 2 mod 4 = 1 → a ≤ b → b ≠ a
45 23 44 biimtrdi ⊢ a ∈ ℕ → 2 ⁢ a 2 ∈ ℙ → 2 ⁢ a 2 mod 4 = 1 → a ≤ b → b ≠ a
46 45 impcomd ⊢ a ∈ ℕ → 2 ⁢ a 2 mod 4 = 1 ∧ 2 ⁢ a 2 ∈ ℙ → a ≤ b → b ≠ a
47 46 com12 ⊢ 2 ⁢ a 2 mod 4 = 1 ∧ 2 ⁢ a 2 ∈ ℙ → a ∈ ℕ → a ≤ b → b ≠ a
48 17 47 biimtrdi ⊢ P = 2 ⁢ a 2 → P mod 4 = 1 ∧ P ∈ ℙ → a ∈ ℕ → a ≤ b → b ≠ a
49 48 expd ⊢ P = 2 ⁢ a 2 → P mod 4 = 1 → P ∈ ℙ → a ∈ ℕ → a ≤ b → b ≠ a
50 49 com34 ⊢ P = 2 ⁢ a 2 → P mod 4 = 1 → a ∈ ℕ → P ∈ ℙ → a ≤ b → b ≠ a
51 50 eqcoms ⊢ 2 ⁢ a 2 = P → P mod 4 = 1 → a ∈ ℕ → P ∈ ℙ → a ≤ b → b ≠ a
52 51 com14 ⊢ P ∈ ℙ → P mod 4 = 1 → a ∈ ℕ → 2 ⁢ a 2 = P → a ≤ b → b ≠ a
53 52 imp31 ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ → 2 ⁢ a 2 = P → a ≤ b → b ≠ a
54 53 ad2antrl ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → 2 ⁢ a 2 = P → a ≤ b → b ≠ a
55 13 54 sylbid ⊢ b = a ∧ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + b 2 = P → a ≤ b → b ≠ a
56 55 expimpd ⊢ b = a → P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b → b ≠ a
57 2a1 ⊢ b ≠ a → P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b → b ≠ a
58 56 57 pm2.61ine ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b → b ≠ a
59 58 pm4.71d ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b ↔ a ≤ b ∧ b ≠ a
60 nnre ⊢ a ∈ ℕ → a ∈ ℝ
61 60 adantl ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ → a ∈ ℝ
62 nnre ⊢ b ∈ ℕ → b ∈ ℝ
63 ltlen ⊢ a ∈ ℝ ∧ b ∈ ℝ → a < b ↔ a ≤ b ∧ b ≠ a
64 61 62 63 syl2an ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a < b ↔ a ≤ b ∧ b ≠ a
65 64 bibi2d ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a ≤ b ↔ a < b ↔ a ≤ b ↔ a ≤ b ∧ b ≠ a
66 65 adantr ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b ↔ a < b ↔ a ≤ b ↔ a ≤ b ∧ b ≠ a
67 59 66 mpbird ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ ∧ a 2 + b 2 = P → a ≤ b ↔ a < b
68 67 ex ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a 2 + b 2 = P → a ≤ b ↔ a < b
69 68 pm5.32rd ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ ∧ b ∈ ℕ → a ≤ b ∧ a 2 + b 2 = P ↔ a < b ∧ a 2 + b 2 = P
70 69 reubidva ⊢ P ∈ ℙ ∧ P mod 4 = 1 ∧ a ∈ ℕ → ∃! b ∈ ℕ a ≤ b ∧ a 2 + b 2 = P ↔ ∃! b ∈ ℕ a < b ∧ a 2 + b 2 = P
71 70 reubidva ⊢ P ∈ ℙ ∧ P mod 4 = 1 → ∃! a ∈ ℕ ∃! b ∈ ℕ a ≤ b ∧ a 2 + b 2 = P ↔ ∃! a ∈ ℕ ∃! b ∈ ℕ a < b ∧ a 2 + b 2 = P
72 1 71 mpbid ⊢ P ∈ ℙ ∧ P mod 4 = 1 → ∃! a ∈ ℕ ∃! b ∈ ℕ a < b ∧ a 2 + b 2 = P