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Description: Lemma for pythagtrip . Show that ( sqrt( C + B ) ) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | pythagtriplem9 | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 + 𝐵 ) ) ∈ ℕ ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pythagtriplem7 | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 + 𝐵 ) ) = ( ( 𝐶 + 𝐵 ) gcd 𝐴 ) ) | |
| 2 | nnz | ⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℤ ) | |
| 3 | nnz | ⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℤ ) | |
| 4 | zaddcl | ⊢ ( ( 𝐶 ∈ ℤ ∧ 𝐵 ∈ ℤ ) → ( 𝐶 + 𝐵 ) ∈ ℤ ) | |
| 5 | 2 3 4 | syl2anr | ⊢ ( ( 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐵 ) ∈ ℤ ) |
| 6 | 5 | 3adant1 | ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐵 ) ∈ ℤ ) |
| 7 | 6 | 3ad2ant1 | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 + 𝐵 ) ∈ ℤ ) |
| 8 | nnz | ⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℤ ) | |
| 9 | 8 | 3ad2ant1 | ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℤ ) |
| 10 | 9 | 3ad2ant1 | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 ∈ ℤ ) |
| 11 | nnne0 | ⊢ ( 𝐴 ∈ ℕ → 𝐴 ≠ 0 ) | |
| 12 | 11 | neneqd | ⊢ ( 𝐴 ∈ ℕ → ¬ 𝐴 = 0 ) |
| 13 | 12 | intnand | ⊢ ( 𝐴 ∈ ℕ → ¬ ( ( 𝐶 + 𝐵 ) = 0 ∧ 𝐴 = 0 ) ) |
| 14 | 13 | 3ad2ant1 | ⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ¬ ( ( 𝐶 + 𝐵 ) = 0 ∧ 𝐴 = 0 ) ) |
| 15 | 14 | 3ad2ant1 | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ¬ ( ( 𝐶 + 𝐵 ) = 0 ∧ 𝐴 = 0 ) ) |
| 16 | gcdn0cl | ⊢ ( ( ( ( 𝐶 + 𝐵 ) ∈ ℤ ∧ 𝐴 ∈ ℤ ) ∧ ¬ ( ( 𝐶 + 𝐵 ) = 0 ∧ 𝐴 = 0 ) ) → ( ( 𝐶 + 𝐵 ) gcd 𝐴 ) ∈ ℕ ) | |
| 17 | 7 10 15 16 | syl21anc | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 + 𝐵 ) gcd 𝐴 ) ∈ ℕ ) |
| 18 | 1 17 | eqeltrd | ⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 + 𝐵 ) ) ∈ ℕ ) |