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:
  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
  have h4 : 0 < 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
htt : PythagoreanTriple a n m
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
⊢ 0 < m

Next tactic:
apply lt_of_le_of_ne ht6
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    

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
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
⊢ 0 ≠ m

Next tactic:
rintro rfl
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    

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
n : ℤ
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
ht1 : a ^ 2 = 0 ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * 0 * n
ht3 : c = 0 ^ 2 + n ^ 2
ht4 : Int.gcd 0 n = 1
ht5 : 0 % 2 = 0 ∧ n % 2 = 1 ∨ 0 % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ 0
htt : PythagoreanTriple a n 0
⊢ False

Next tactic:
revert hb20
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    

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
n : ℤ
h3 : Int.gcd a n = 1
ht1 : a ^ 2 = 0 ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * 0 * n
ht3 : c = 0 ^ 2 + n ^ 2
ht4 : Int.gcd 0 n = 1
ht5 : 0 % 2 = 0 ∧ n % 2 = 1 ∨ 0 % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ 0
htt : PythagoreanTriple a n 0
⊢ b ^ 2 ≠ 0 → False

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

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
n : ℤ
h3 : Int.gcd a n = 1
ht1 : a ^ 2 = 0 ^ 2 - n ^ 2
ht2 : b ^ 2 = 2 * 0 * n
ht3 : c = 0 ^ 2 + n ^ 2
ht4 : Int.gcd 0 n = 1
ht5 : 0 % 2 = 0 ∧ n % 2 = 1 ∨ 0 % 2 = 1 ∧ n % 2 = 0
ht6 : 0 ≤ 0
htt : PythagoreanTriple a n 0
⊢ 2 * 0 * n ≠ 0 → False

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

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
h4 : 0 < m
⊢ False

Next tactic:
obtain ⟨r, s, _, htt2, htt3, htt4, htt5, htt6⟩ := htt.coprime_classification' h3 ha2 h4
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    rw [ht2]
    simp
  obtain ⟨r, s, _, htt2, htt3, htt4, htt5, htt6⟩ := htt.coprime_classification' h3 ha2 h4
  -- Now use the fact that (b / 2) ^ 2 = m * r * s, and m, r and s are pairwise coprime to obtain
  -- i, j and k such that m = i ^ 2, r = j ^ 2 and s = k ^ 2.
  -- m and r * s are coprime because m = r ^ 2 + s ^ 2 and r and s are coprime.
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro.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
h4 : 0 < m
r s : ℤ
left✝ : a = r ^ 2 - s ^ 2
htt2 : n = 2 * r * s
htt3 : m = r ^ 2 + s ^ 2
htt4 : Int.gcd r s = 1
htt5 : r % 2 = 0 ∧ s % 2 = 1 ∨ r % 2 = 1 ∧ s % 2 = 0
htt6 : 0 ≤ r
⊢ False

Next tactic:
have hcp : Int.gcd m (r * s) = 1 := by
    rw [htt3]
    exact
      Int.gcd_eq_one_iff_coprime.mpr (Int.coprime_of_sq_sum' (Int.gcd_eq_one_iff_coprime.mp htt4))
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    rw [ht2]
    simp
  obtain ⟨r, s, _, htt2, htt3, htt4, htt5, htt6⟩ := htt.coprime_classification' h3 ha2 h4
  -- Now use the fact that (b / 2) ^ 2 = m * r * s, and m, r and s are pairwise coprime to obtain
  -- i, j and k such that m = i ^ 2, r = j ^ 2 and s = k ^ 2.
  -- m and r * s are coprime because m = r ^ 2 + s ^ 2 and r and s are coprime.
  have hcp : Int.gcd m (r * s) = 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
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
h4 : 0 < m
r s : ℤ
left✝ : a = r ^ 2 - s ^ 2
htt2 : n = 2 * r * s
htt3 : m = r ^ 2 + s ^ 2
htt4 : Int.gcd r s = 1
htt5 : r % 2 = 0 ∧ s % 2 = 1 ∨ r % 2 = 1 ∧ s % 2 = 0
htt6 : 0 ≤ r
⊢ Int.gcd m (r * s) = 1

Next tactic:
rw [htt3]
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    rw [ht2]
    simp
  obtain ⟨r, s, _, htt2, htt3, htt4, htt5, htt6⟩ := htt.coprime_classification' h3 ha2 h4
  -- Now use the fact that (b / 2) ^ 2 = m * r * s, and m, r and s are pairwise coprime to obtain
  -- i, j and k such that m = i ^ 2, r = j ^ 2 and s = k ^ 2.
  -- m and r * s are coprime because m = r ^ 2 + s ^ 2 and r and s are coprime.
  have hcp : Int.gcd m (r * s) = 1 := by
    rw [htt3]
    

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
h3 : Int.gcd a n = 1
hb20 : b ^ 2 ≠ 0
h4 : 0 < m
r s : ℤ
left✝ : a = r ^ 2 - s ^ 2
htt2 : n = 2 * r * s
htt3 : m = r ^ 2 + s ^ 2
htt4 : Int.gcd r s = 1
htt5 : r % 2 = 0 ∧ s % 2 = 1 ∨ r % 2 = 1 ∧ s % 2 = 0
htt6 : 0 ≤ r
⊢ Int.gcd (r ^ 2 + s ^ 2) (r * s) = 1

Next tactic:
exact
      Int.gcd_eq_one_iff_coprime.mpr (Int.coprime_of_sq_sum' (Int.gcd_eq_one_iff_coprime.mp htt4))
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
  have h4 : 0 < m := by
    apply lt_of_le_of_ne ht6
    rintro rfl
    revert hb20
    rw [ht2]
    simp
  obtain ⟨r, s, _, htt2, htt3, htt4, htt5, htt6⟩ := htt.coprime_classification' h3 ha2 h4
  -- Now use the fact that (b / 2) ^ 2 = m * r * s, and m, r and s are pairwise coprime to obtain
  -- i, j and k such that m = i ^ 2, r = j ^ 2 and s = k ^ 2.
  -- m and r * s are coprime because m = r ^ 2 + s ^ 2 and r and s are coprime.
  have hcp : Int.gcd m (r * s) = 1 := by
    rw [htt3]
    exact
      Int.gcd_eq_one_iff_coprime.mpr (Int.coprime_of_sq_sum' (Int.gcd_eq_one_iff_coprime.mp htt4))
  -- b is even because b ^ 2 = 2 * m * n.
  

Proof state:
case intro.intro.intro.intro.intro.intro.intro.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
h4 : 0 < m
r s : ℤ
left✝ : a = r ^ 2 - s ^ 2
htt2 : n = 2 * r * s
htt3 : m = r ^ 2 + s ^ 2
htt4 : Int.gcd r s = 1
htt5 : r % 2 = 0 ∧ s % 2 = 1 ∨ r % 2 = 1 ∧ s % 2 = 0
htt6 : 0 ≤ r
hcp : Int.gcd m (r * s) = 1
⊢ False

Next tactic:
have hb2 : 2 ∣ b := by
    apply @Int.Prime.dvd_pow' _ 2 _ Nat.prime_two
    rw [ht2, mul_assoc]
    exact dvd_mul_right 2 (m * n)