Metamath Proof Explorer


Theorem 4sqlem4

Description: Lemma for 4sq . We can express the four-square property more compactly in terms of gaussian integers, because the norms of gaussian integers are exactly sums of two squares. (Contributed by Mario Carneiro, 14-Jul-2014)

Ref Expression
Hypothesis 4sq.1 ⊢ S = n | ∃ x ∈ ℤ ∃ y ∈ ℤ ∃ z ∈ ℤ ∃ w ∈ ℤ n = x 2 + y 2 + z 2 + w 2
Assertion 4sqlem4 ⊢ A ∈ S ↔ ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2

Proof

Step Hyp Ref Expression
1 4sq.1 ⊢ S = n | ∃ x ∈ ℤ ∃ y ∈ ℤ ∃ z ∈ ℤ ∃ w ∈ ℤ n = x 2 + y 2 + z 2 + w 2
2 1 4sqlem2 ⊢ A ∈ S ↔ ∃ a ∈ ℤ ∃ b ∈ ℤ ∃ c ∈ ℤ ∃ d ∈ ℤ A = a 2 + b 2 + c 2 + d 2
3 gzreim ⊢ a ∈ ℤ ∧ b ∈ ℤ → a + i ⁢ b ∈ ℤ i
4 3 adantr ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → a + i ⁢ b ∈ ℤ i
5 gzreim ⊢ c ∈ ℤ ∧ d ∈ ℤ → c + i ⁢ d ∈ ℤ i
6 5 adantl ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → c + i ⁢ d ∈ ℤ i
7 gzcn ⊢ a + i ⁢ b ∈ ℤ i → a + i ⁢ b ∈ ℂ
8 3 7 syl ⊢ a ∈ ℤ ∧ b ∈ ℤ → a + i ⁢ b ∈ ℂ
9 8 absvalsq2d ⊢ a ∈ ℤ ∧ b ∈ ℤ → a + i ⁢ b 2 = ℜ ⁡ a + i ⁢ b 2 + ℑ ⁡ a + i ⁢ b 2
10 zre ⊢ a ∈ ℤ → a ∈ ℝ
11 zre ⊢ b ∈ ℤ → b ∈ ℝ
12 crre ⊢ a ∈ ℝ ∧ b ∈ ℝ → ℜ ⁡ a + i ⁢ b = a
13 10 11 12 syl2an ⊢ a ∈ ℤ ∧ b ∈ ℤ → ℜ ⁡ a + i ⁢ b = a
14 13 oveq1d ⊢ a ∈ ℤ ∧ b ∈ ℤ → ℜ ⁡ a + i ⁢ b 2 = a 2
15 crim ⊢ a ∈ ℝ ∧ b ∈ ℝ → ℑ ⁡ a + i ⁢ b = b
16 10 11 15 syl2an ⊢ a ∈ ℤ ∧ b ∈ ℤ → ℑ ⁡ a + i ⁢ b = b
17 16 oveq1d ⊢ a ∈ ℤ ∧ b ∈ ℤ → ℑ ⁡ a + i ⁢ b 2 = b 2
18 14 17 oveq12d ⊢ a ∈ ℤ ∧ b ∈ ℤ → ℜ ⁡ a + i ⁢ b 2 + ℑ ⁡ a + i ⁢ b 2 = a 2 + b 2
19 9 18 eqtrd ⊢ a ∈ ℤ ∧ b ∈ ℤ → a + i ⁢ b 2 = a 2 + b 2
20 gzcn ⊢ c + i ⁢ d ∈ ℤ i → c + i ⁢ d ∈ ℂ
21 5 20 syl ⊢ c ∈ ℤ ∧ d ∈ ℤ → c + i ⁢ d ∈ ℂ
22 21 absvalsq2d ⊢ c ∈ ℤ ∧ d ∈ ℤ → c + i ⁢ d 2 = ℜ ⁡ c + i ⁢ d 2 + ℑ ⁡ c + i ⁢ d 2
23 zre ⊢ c ∈ ℤ → c ∈ ℝ
24 zre ⊢ d ∈ ℤ → d ∈ ℝ
25 crre ⊢ c ∈ ℝ ∧ d ∈ ℝ → ℜ ⁡ c + i ⁢ d = c
26 23 24 25 syl2an ⊢ c ∈ ℤ ∧ d ∈ ℤ → ℜ ⁡ c + i ⁢ d = c
27 26 oveq1d ⊢ c ∈ ℤ ∧ d ∈ ℤ → ℜ ⁡ c + i ⁢ d 2 = c 2
28 crim ⊢ c ∈ ℝ ∧ d ∈ ℝ → ℑ ⁡ c + i ⁢ d = d
29 23 24 28 syl2an ⊢ c ∈ ℤ ∧ d ∈ ℤ → ℑ ⁡ c + i ⁢ d = d
30 29 oveq1d ⊢ c ∈ ℤ ∧ d ∈ ℤ → ℑ ⁡ c + i ⁢ d 2 = d 2
31 27 30 oveq12d ⊢ c ∈ ℤ ∧ d ∈ ℤ → ℜ ⁡ c + i ⁢ d 2 + ℑ ⁡ c + i ⁢ d 2 = c 2 + d 2
32 22 31 eqtrd ⊢ c ∈ ℤ ∧ d ∈ ℤ → c + i ⁢ d 2 = c 2 + d 2
33 19 32 oveqan12d ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → a + i ⁢ b 2 + c + i ⁢ d 2 = a 2 + b 2 + c 2 + d 2
34 33 eqcomd ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → a 2 + b 2 + c 2 + d 2 = a + i ⁢ b 2 + c + i ⁢ d 2
35 fveq2 ⊢ u = a + i ⁢ b → u = a + i ⁢ b
36 35 oveq1d ⊢ u = a + i ⁢ b → u 2 = a + i ⁢ b 2
37 36 oveq1d ⊢ u = a + i ⁢ b → u 2 + v 2 = a + i ⁢ b 2 + v 2
38 37 eqeq2d ⊢ u = a + i ⁢ b → a 2 + b 2 + c 2 + d 2 = u 2 + v 2 ↔ a 2 + b 2 + c 2 + d 2 = a + i ⁢ b 2 + v 2
39 fveq2 ⊢ v = c + i ⁢ d → v = c + i ⁢ d
40 39 oveq1d ⊢ v = c + i ⁢ d → v 2 = c + i ⁢ d 2
41 40 oveq2d ⊢ v = c + i ⁢ d → a + i ⁢ b 2 + v 2 = a + i ⁢ b 2 + c + i ⁢ d 2
42 41 eqeq2d ⊢ v = c + i ⁢ d → a 2 + b 2 + c 2 + d 2 = a + i ⁢ b 2 + v 2 ↔ a 2 + b 2 + c 2 + d 2 = a + i ⁢ b 2 + c + i ⁢ d 2
43 38 42 rspc2ev ⊢ a + i ⁢ b ∈ ℤ i ∧ c + i ⁢ d ∈ ℤ i ∧ a 2 + b 2 + c 2 + d 2 = a + i ⁢ b 2 + c + i ⁢ d 2 → ∃ u ∈ ℤ i ∃ v ∈ ℤ i a 2 + b 2 + c 2 + d 2 = u 2 + v 2
44 4 6 34 43 syl3anc ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → ∃ u ∈ ℤ i ∃ v ∈ ℤ i a 2 + b 2 + c 2 + d 2 = u 2 + v 2
45 eqeq1 ⊢ A = a 2 + b 2 + c 2 + d 2 → A = u 2 + v 2 ↔ a 2 + b 2 + c 2 + d 2 = u 2 + v 2
46 45 2rexbidv ⊢ A = a 2 + b 2 + c 2 + d 2 → ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2 ↔ ∃ u ∈ ℤ i ∃ v ∈ ℤ i a 2 + b 2 + c 2 + d 2 = u 2 + v 2
47 44 46 syl5ibrcom ⊢ a ∈ ℤ ∧ b ∈ ℤ ∧ c ∈ ℤ ∧ d ∈ ℤ → A = a 2 + b 2 + c 2 + d 2 → ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2
48 47 rexlimdvva ⊢ a ∈ ℤ ∧ b ∈ ℤ → ∃ c ∈ ℤ ∃ d ∈ ℤ A = a 2 + b 2 + c 2 + d 2 → ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2
49 48 rexlimivv ⊢ ∃ a ∈ ℤ ∃ b ∈ ℤ ∃ c ∈ ℤ ∃ d ∈ ℤ A = a 2 + b 2 + c 2 + d 2 → ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2
50 2 49 sylbi ⊢ A ∈ S → ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2
51 1 4sqlem4a ⊢ u ∈ ℤ i ∧ v ∈ ℤ i → u 2 + v 2 ∈ S
52 eleq1a ⊢ u 2 + v 2 ∈ S → A = u 2 + v 2 → A ∈ S
53 51 52 syl ⊢ u ∈ ℤ i ∧ v ∈ ℤ i → A = u 2 + v 2 → A ∈ S
54 53 rexlimivv ⊢ ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2 → A ∈ S
55 50 54 impbii ⊢ A ∈ S ↔ ∃ u ∈ ℤ i ∃ v ∈ ℤ i A = u 2 + v 2