Metamath Proof Explorer


Theorem 2sqn0

Description: If the sum of two squares is prime, none of the original number is zero. (Contributed by Thierry Arnoux, 4-Feb-2020)

Ref Expression
Hypotheses 2sqcoprm.1 ⊢ φ → P ∈ ℙ
2sqcoprm.2 ⊢ φ → A ∈ ℤ
2sqcoprm.3 ⊢ φ → B ∈ ℤ
2sqcoprm.4 ⊢ φ → A 2 + B 2 = P
Assertion 2sqn0 ⊢ φ → A ≠ 0

Proof

Step Hyp Ref Expression
1 2sqcoprm.1 ⊢ φ → P ∈ ℙ
2 2sqcoprm.2 ⊢ φ → A ∈ ℤ
3 2sqcoprm.3 ⊢ φ → B ∈ ℤ
4 2sqcoprm.4 ⊢ φ → A 2 + B 2 = P
5 4 1 eqeltrd ⊢ φ → A 2 + B 2 ∈ ℙ
6 5 adantr ⊢ φ ∧ A = 0 → A 2 + B 2 ∈ ℙ
7 sq0i ⊢ A = 0 → A 2 = 0
8 7 oveq1d ⊢ A = 0 → A 2 + B 2 = 0 + B 2
9 3 zcnd ⊢ φ → B ∈ ℂ
10 9 sqcld ⊢ φ → B 2 ∈ ℂ
11 10 addlidd ⊢ φ → 0 + B 2 = B 2
12 8 11 sylan9eqr ⊢ φ ∧ A = 0 → A 2 + B 2 = B 2
13 sqnprm ⊢ B ∈ ℤ → ¬ B 2 ∈ ℙ
14 3 13 syl ⊢ φ → ¬ B 2 ∈ ℙ
15 14 adantr ⊢ φ ∧ A = 0 → ¬ B 2 ∈ ℙ
16 12 15 eqneltrd ⊢ φ ∧ A = 0 → ¬ A 2 + B 2 ∈ ℙ
17 6 16 pm2.65da ⊢ φ → ¬ A = 0
18 17 neqned ⊢ φ → A ≠ 0