Metamath Proof Explorer


Theorem fltmul

Description: A counterexample to FLT stays valid when scaled. The hypotheses are more general than they need to be for convenience. (There does not seem to be a standard term for Fermat or Pythagorean triples extended to any N e. NN0 , so the label is more about the context in which this theorem is used). (Contributed by SN, 20-Aug-2024)

Ref Expression
Hypotheses fltmul.s ⊢ φ → S ∈ ℂ
fltmul.a ⊢ φ → A ∈ ℂ
fltmul.b ⊢ φ → B ∈ ℂ
fltmul.c ⊢ φ → C ∈ ℂ
fltmul.n ⊢ φ → N ∈ ℕ 0
fltmul.1 ⊢ φ → A N + B N = C N
Assertion fltmul ⊢ φ → S ⁢ A N + S ⁢ B N = S ⁢ C N

Proof

Step Hyp Ref Expression
1 fltmul.s ⊢ φ → S ∈ ℂ
2 fltmul.a ⊢ φ → A ∈ ℂ
3 fltmul.b ⊢ φ → B ∈ ℂ
4 fltmul.c ⊢ φ → C ∈ ℂ
5 fltmul.n ⊢ φ → N ∈ ℕ 0
6 fltmul.1 ⊢ φ → A N + B N = C N
7 1 5 expcld ⊢ φ → S N ∈ ℂ
8 2 5 expcld ⊢ φ → A N ∈ ℂ
9 3 5 expcld ⊢ φ → B N ∈ ℂ
10 7 8 9 adddid ⊢ φ → S N ⁢ A N + B N = S N ⁢ A N + S N ⁢ B N
11 6 oveq2d ⊢ φ → S N ⁢ A N + B N = S N ⁢ C N
12 10 11 eqtr3d ⊢ φ → S N ⁢ A N + S N ⁢ B N = S N ⁢ C N
13 1 2 5 mulexpd ⊢ φ → S ⁢ A N = S N ⁢ A N
14 1 3 5 mulexpd ⊢ φ → S ⁢ B N = S N ⁢ B N
15 13 14 oveq12d ⊢ φ → S ⁢ A N + S ⁢ B N = S N ⁢ A N + S N ⁢ B N
16 1 4 5 mulexpd ⊢ φ → S ⁢ C N = S N ⁢ C N
17 12 15 16 3eqtr4d ⊢ φ → S ⁢ A N + S ⁢ B N = S ⁢ C N