l3lab/ntp-mathlib huggingface dataset:
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
⊢ False

Next tactic:
have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
⊢ False

Next tactic:
have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
⊢ a ^ 2 % 2 = 1

Next tactic:
rw [sq, Int.mul_emod, ha2]
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
⊢ 1 * 1 % 2 = 1

Next tactic:
decide
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
⊢ False

Next tactic:
obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
⊢ False

Next tactic:
have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
⊢ PythagoreanTriple a n m

Next tactic:
delta PythagoreanTriple
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
⊢ a * a + n * n = m * m

Next tactic:
linear_combination ht1
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ False

Next tactic:
have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    rw [← sq, ht1, (by ring : m ^ 2 - n ^ 2 = m ^ 2 + -n * n)]
    exact (Int.gcd_eq_one_iff_coprime.mp ht4).pow_left.add_mul_right_left (-n)
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ Int.gcd a n = 1

Next tactic:
apply Int.gcd_eq_one_iff_coprime.mpr
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ IsCoprime a n

Next tactic:
apply @IsCoprime.of_mul_left_left _ _ _ a
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    

Proof state:
case H
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ IsCoprime (a * a) n

Next tactic:
rw [← sq, ht1, (by ring : m ^ 2 - n ^ 2 = m ^ 2 + -n * n)]
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    rw [← sq, ht1, (by 

Proof state:
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ m ^ 2 - n ^ 2 = m ^ 2 + -n * n

Next tactic:
ring
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    rw [← sq, ht1, (by ring : m ^ 2 - n ^ 2 = m ^ 2 + -n * n)]
    

Proof state:
case H
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
⊢ IsCoprime (m ^ 2 + -n * n) n

Next tactic:
exact (Int.gcd_eq_one_iff_coprime.mp ht4).pow_left.add_mul_right_left (-n)
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    rw [← sq, ht1, (by ring : m ^ 2 - n ^ 2 = m ^ 2 + -n * n)]
    exact (Int.gcd_eq_one_iff_coprime.mp ht4).pow_left.add_mul_right_left (-n)
  -- m is positive because b is non-zero and b ^ 2 = 2 * m * n and we already have 0 ≤ m.
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
h3 : Int.gcd a n = 1
⊢ False

Next tactic:
have hb20 : b ^ 2 ≠ 0 := mt pow_eq_zero h.1.2.1
theorem not_minimal {a b c : ℤ} (h : Minimal a b c) (ha2 : a % 2 = 1) (hc : 0 < c) : False := by
  -- Use the fact that a ^ 2, b ^ 2, c form a pythagorean triple to obtain m and n such that
  -- a ^ 2 = m ^ 2 - n ^ 2, b ^ 2 = 2 * m * n and c = m ^ 2 + n ^ 2
  -- first the formula:
  have ht : PythagoreanTriple (a ^ 2) (b ^ 2) c := by
    delta PythagoreanTriple
    linear_combination h.1.2.2
  -- coprime requirement:
  have h2 : Int.gcd (a ^ 2) (b ^ 2) = 1 := Int.gcd_eq_one_iff_coprime.mpr (coprime_of_minimal h).pow
  -- in order to reduce the possibilities we get from the classification of pythagorean triples
  -- it helps if we know the parity of a ^ 2 (and the sign of c):
  have ha22 : a ^ 2 % 2 = 1 := by
    rw [sq, Int.mul_emod, ha2]
    decide
  obtain ⟨m, n, ht1, ht2, ht3, ht4, ht5, ht6⟩ := ht.coprime_classification' h2 ha22 hc
  -- Now a, n, m form a pythagorean triple and so we can obtain r and s such that
  -- a = r ^ 2 - s ^ 2, n = 2 * r * s and m = r ^ 2 + s ^ 2
  -- formula:
  have htt : PythagoreanTriple a n m := by
    delta PythagoreanTriple
    linear_combination ht1
  -- a and n are coprime, because a ^ 2 = m ^ 2 - n ^ 2 and m and n are coprime.
  have h3 : Int.gcd a n = 1 := by
    apply Int.gcd_eq_one_iff_coprime.mpr
    apply @IsCoprime.of_mul_left_left _ _ _ a
    rw [← sq, ht1, (by ring : m ^ 2 - n ^ 2 = m ^ 2 + -n * n)]
    exact (Int.gcd_eq_one_iff_coprime.mp ht4).pow_left.add_mul_right_left (-n)
  -- m is positive because b is non-zero and b ^ 2 = 2 * m * n and we already have 0 ≤ m.
  have hb20 : b ^ 2 ≠ 0 := mt pow_eq_zero h.1.2.1
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro
a b c : ℤ
h : Minimal a b c
ha2 : a % 2 = 1
hc : 0 < c
ht : PythagoreanTriple (a ^ 2) (b ^ 2) c
h2 : Int.gcd (a ^ 2) (b ^ 2) = 1
ha22 : a ^ 2 % 2 = 1
m n : ℤ
ht1 : a ^ 2 = m ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * m * n
ht3 : c = m ^ 2 + n ^ 2
ht4 : Int.gcd m n = 1
ht5 : m % 2 = 0 ∧ n % 2 = 1 ∨ m % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ m
htt : PythagoreanTriple a n m
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
⊢ False

Next tactic:
have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    rw [ht2]
    simp