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Metamath Proof Explorer

Theorem pythagtriplem15

Description: Lemma for pythagtrip . Show the relationship between M , N , and A . (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 pythagtriplem15 ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 = ( ( 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 simp3 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℕ )
7 simp1 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℕ )
8 6 7 nnaddcld ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐴 ) ∈ ℕ )
9 8 nncnd ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐴 ) ∈ ℂ )
10 9 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 + 𝐴 ) ∈ ℂ )
11 nnz ⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℤ )
12 11 3ad2ant3 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℤ )
13 nnz ⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℤ )
14 13 3ad2ant1 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℤ )
15 12 14 zsubcld ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 − 𝐴 ) ∈ ℤ )
16 15 zcnd ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
17 16 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐴 ) ∈ ℂ )
18 2cnne0 ⊢ ( 2 ∈ ℂ ∧ 2 ≠ 0 )
19 divsubdir ⊢ ( ( ( 𝐶 + 𝐴 ) ∈ ℂ ∧ ( 𝐶 − 𝐴 ) ∈ ℂ ∧ ( 2 ∈ ℂ ∧ 2 ≠ 0 ) ) → ( ( ( 𝐶 + 𝐴 ) − ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) − ( ( 𝐶 − 𝐴 ) / 2 ) ) )
20 18 19 mp3an3 ⊢ ( ( ( 𝐶 + 𝐴 ) ∈ ℂ ∧ ( 𝐶 − 𝐴 ) ∈ ℂ ) → ( ( ( 𝐶 + 𝐴 ) − ( 𝐶 − 𝐴 ) ) / 2 ) = ( ( ( 𝐶 + 𝐴 ) / 2 ) − ( ( 𝐶 − 𝐴 ) / 2 ) ) )
21 10 17 20 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 ) ) )
22 5 21 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 ) )
23 nncn ⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℂ )
24 23 3ad2ant3 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℂ )
25 24 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐶 ∈ ℂ )
26 nncn ⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℂ )
27 26 3ad2ant1 ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℂ )
28 27 3ad2ant1 ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 ∈ ℂ )
29 25 28 28 pnncand ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 − 𝐴 ) ) = ( 𝐴 + 𝐴 ) )
30 28 2timesd ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 2 · 𝐴 ) = ( 𝐴 + 𝐴 ) )
31 29 30 eqtr4d ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 − 𝐴 ) ) = ( 2 · 𝐴 ) )
32 31 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 · 𝐴 ) / 2 ) )
33 2cn ⊢ 2 ∈ ℂ
34 2ne0 ⊢ 2 ≠ 0
35 divcan3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ) → ( ( 2 · 𝐴 ) / 2 ) = 𝐴 )
36 33 34 35 mp3an23 ⊢ ( 𝐴 ∈ ℂ → ( ( 2 · 𝐴 ) / 2 ) = 𝐴 )
37 28 36 syl ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 2 · 𝐴 ) / 2 ) = 𝐴 )
38 22 32 37 3eqtrrd ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 = ( ( 𝑀 ↑ 2 ) − ( 𝑁 ↑ 2 ) ) )