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Metamath Proof Explorer

Theorem pythagtriplem17

Description: Lemma for pythagtrip . Show the relationship between M , N , and C . (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Hypotheses pythagtriplem15.1 ⊢ 𝑀 = ( ( ( √ ‘ ( 𝐶 + 𝐵 ) ) + ( √ ‘ ( 𝐶 − 𝐵 ) ) ) / 2 )
pythagtriplem15.2 ⊢ 𝑁 = ( ( ( √ ‘ ( 𝐶 + 𝐵 ) ) − ( √ ‘ ( 𝐶 − 𝐵 ) ) ) / 2 )
Assertion pythagtriplem17 ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐶 = ( ( 𝑀 ↑ 2 ) + ( 𝑁 ↑ 2 ) ) )

Proof

Step Hyp Ref Expression
1 pythagtriplem15.1 ⊢ 𝑀 = ( ( ( √ ‘ ( 𝐶 + 𝐵 ) ) + ( √ ‘ ( 𝐶 − 𝐵 ) ) ) / 2 )
2 pythagtriplem15.2 ⊢ 𝑁 = ( ( ( √ ‘ ( 𝐶 + 𝐵 ) ) − ( √ ‘ ( 𝐶 − 𝐵 ) ) ) / 2 )
3 1 pythagtriplem12 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝑀 ↑ 2 ) = ( ( 𝐶 + 𝐴 ) / 2 ) )
4 2 pythagtriplem14 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝑁 ↑ 2 ) = ( ( 𝐶 − 𝐴 ) / 2 ) )
5 3 4 oveq12d ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝑀 ↑ 2 ) + ( 𝑁 ↑ 2 ) ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) + ( ( 𝐶 − 𝐴 ) / 2 ) ) )
6 nncn ⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℂ )
7 6 3ad2ant3 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℂ )
8 7 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐶 ∈ ℂ )
9 nncn ⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℂ )
10 9 3ad2ant1 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℂ )
11 10 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 ∈ ℂ )
12 8 11 addcld ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 + 𝐴 ) ∈ ℂ )
13 8 11 subcld ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
14 2cnne0 ⊢ ( 2 ∈ ℂ ∧ 2 ≠ 0 )
15 divdir ⊢ ( ( ( 𝐶 + 𝐴 ) ∈ ℂ ∧ ( 𝐶 − 𝐴 ) ∈ ℂ ∧ ( 2 ∈ ℂ ∧ 2 ≠ 0 ) ) → ( ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) + ( ( 𝐶 − 𝐴 ) / 2 ) ) )
16 14 15 mp3an3 ⊢ ( ( ( 𝐶 + 𝐴 ) ∈ ℂ ∧ ( 𝐶 − 𝐴 ) ∈ ℂ ) → ( ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) + ( ( 𝐶 − 𝐴 ) / 2 ) ) )
17 12 13 16 syl2anc ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) + ( ( 𝐶 − 𝐴 ) / 2 ) ) )
18 5 17 eqtr4d ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝑀 ↑ 2 ) + ( 𝑁 ↑ 2 ) ) = ( ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) / 2 ) )
19 8 11 8 ppncand ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) = ( 𝐶 + 𝐶 ) )
20 8 2timesd ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 2 · 𝐶 ) = ( 𝐶 + 𝐶 ) )
21 19 20 eqtr4d ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) = ( 2 · 𝐶 ) )
22 21 oveq1d ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐶 + 𝐴 ) + ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( 2 · 𝐶 ) / 2 ) )
23 2cn ⊢ 2 ∈ ℂ
24 2ne0 ⊢ 2 ≠ 0
25 divcan3 ⊢ ( ( 𝐶 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ) → ( ( 2 · 𝐶 ) / 2 ) = 𝐶 )
26 23 24 25 mp3an23 ⊢ ( 𝐶 ∈ ℂ → ( ( 2 · 𝐶 ) / 2 ) = 𝐶 )
27 8 26 syl ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 2 · 𝐶 ) / 2 ) = 𝐶 )
28 18 22 27 3eqtrrd ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐶 = ( ( 𝑀 ↑ 2 ) + ( 𝑁 ↑ 2 ) ) )