l3lab/ntp-mathlib huggingface dataset:
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · 

Proof state:
case pos
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    

Proof state:
case pos
b : ℕ
h✝ : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
h : ∃ p, b * (b ^ 2 - 1) + 1 + m ≤ p ∧ Prime p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
cases' h with p hp
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    

Proof state:
case pos.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp : b * (b ^ 2 - 1) + 1 + m ≤ p ∧ Prime p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
cases' hp with hp₁ hp₂
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by 

Proof state:
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
⊢ 4 > 1

Next tactic:
norm_num
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₅ : b * (b ^ 2 - 1) < p := by linarith
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by 

Proof state:
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
⊢ b * (b ^ 2 - 1) < p

Next tactic:
linarith
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
h₆ : ¬p ∣ b * (b ^ 2 - 1)
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₇ : b ≤ b * (b ^ 2 - 1) := Nat.le_mul_of_pos_right h₃
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
    have h₇ : b ≤ b * (b ^ 2 - 1) := Nat.le_mul_of_pos_right h₃
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
h₆ : ¬p ∣ b * (b ^ 2 - 1)
h₇ : b ≤ b * (b ^ 2 - 1)
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₈ : 2 ≤ b * (b ^ 2 - 1) := le_trans b_ge_two h₇
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
    have h₇ : b ≤ b * (b ^ 2 - 1) := Nat.le_mul_of_pos_right h₃
    have h₈ : 2 ≤ b * (b ^ 2 - 1) := le_trans b_ge_two h₇
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
h₆ : ¬p ∣ b * (b ^ 2 - 1)
h₇ : b ≤ b * (b ^ 2 - 1)
h₈ : 2 ≤ b * (b ^ 2 - 1)
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₉ : 2 < p := gt_of_gt_of_ge h₅ h₈
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
    have h₇ : b ≤ b * (b ^ 2 - 1) := Nat.le_mul_of_pos_right h₃
    have h₈ : 2 ≤ b * (b ^ 2 - 1) := le_trans b_ge_two h₇
    have h₉ : 2 < p := gt_of_gt_of_ge h₅ h₈
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
h₆ : ¬p ∣ b * (b ^ 2 - 1)
h₇ : b ≤ b * (b ^ 2 - 1)
h₈ : 2 ≤ b * (b ^ 2 - 1)
h₉ : 2 < p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
have h₁₀ := psp_from_prime_gt_p b_ge_two hp₂ h₉
/-- For all positive bases, there exist infinite **Fermat pseudoprimes** to that base.
Given in this form: for all numbers `b ≥ 1` and `m`, there exists a pseudoprime `n` to base `b` such
that `m ≤ n`. This form is similar to `Nat.exists_infinite_primes`.
-/
theorem exists_infinite_pseudoprimes {b : ℕ} (h : 1 ≤ b) (m : ℕ) :
    ∃ n : ℕ, FermatPsp n b ∧ m ≤ n := by
  by_cases b_ge_two : 2 ≤ b
  -- If `2 ≤ b`, then because there exist infinite prime numbers, there is a prime number p with
  -- `m ≤ p` and `¬p ∣ b*(b^2 - 1)`. We pick a prime number `b*(b^2 - 1) + 1 + m ≤ p` because we
  -- automatically know that `p` is greater than m and that it does not divide `b*(b^2 - 1)`
  -- (because `p` can't divide a number less than `p`).
  -- From `p`, we can use the lemmas we proved earlier to show that
  -- `((b^p - 1)/(b - 1)) * ((b^p + 1)/(b + 1))` is a pseudoprime to base `b`.
  · have h := Nat.exists_infinite_primes (b * (b ^ 2 - 1) + 1 + m)
    cases' h with p hp
    cases' hp with hp₁ hp₂
    have h₁ : 0 < b := pos_of_gt (Nat.succ_le_iff.mp b_ge_two)
    have h₂ : 4 ≤ b ^ 2 := pow_le_pow_left' b_ge_two 2
    have h₃ : 0 < b ^ 2 - 1 := tsub_pos_of_lt (gt_of_ge_of_gt h₂ (by norm_num))
    have h₄ : 0 < b * (b ^ 2 - 1) := mul_pos h₁ h₃
    have h₅ : b * (b ^ 2 - 1) < p := by linarith
    have h₆ : ¬p ∣ b * (b ^ 2 - 1) := Nat.not_dvd_of_pos_of_lt h₄ h₅
    have h₇ : b ≤ b * (b ^ 2 - 1) := Nat.le_mul_of_pos_right h₃
    have h₈ : 2 ≤ b * (b ^ 2 - 1) := le_trans b_ge_two h₇
    have h₉ : 2 < p := gt_of_gt_of_ge h₅ h₈
    have h₁₀ := psp_from_prime_gt_p b_ge_two hp₂ h₉
    

Proof state:
case pos.intro.intro
b : ℕ
h : 1 ≤ b
m : ℕ
b_ge_two : 2 ≤ b
p : ℕ
hp₁ : b * (b ^ 2 - 1) + 1 + m ≤ p
hp₂ : Prime p
h₁ : 0 < b
h₂ : 4 ≤ b ^ 2
h₃ : 0 < b ^ 2 - 1
h₄ : 0 < b * (b ^ 2 - 1)
h₅ : b * (b ^ 2 - 1) < p
h₆ : ¬p ∣ b * (b ^ 2 - 1)
h₇ : b ≤ b * (b ^ 2 - 1)
h₈ : 2 ≤ b * (b ^ 2 - 1)
h₉ : 2 < p
h₁₀ : p < Nat.psp_from_prime b p
⊢ ∃ n, FermatPsp n b ∧ m ≤ n

Next tactic:
use psp_from_prime b p