/-- Lagrange's **four squares theorem** for a prime number. Use `Nat.sum_four_squares` instead. -/
protected theorem Prime.sum_four_squares {p : ℕ} (hp : p.Prime) :
    ∃ a b c d : ℕ, a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = p := by
  have := Fact.mk hp
  -- Find `a`, `b`, `c`, `d`, `0 < m < p` such that `a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = m * p`
  have natAbs_iff {a b c d : ℤ} {k : ℕ} :
      a.natAbs ^ 2 + b.natAbs ^ 2 + c.natAbs ^ 2 + d.natAbs ^ 2 = k ↔
        a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = k := by
    rw [← @Nat.cast_inj ℤ]; push_cast [sq_abs]; rfl
  have hm : ∃ m < p, 0 < m ∧ ∃ a b c d : ℕ, a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = m * p
  · obtain ⟨a, b, k, hk₀, hkp, hk⟩ := exists_sq_add_sq_add_one_eq_mul p
    refine ⟨k, hkp, hk₀, a, b, 1, 0, ?_⟩
    simpa
  -- Take the minimal possible `m`
  rcases Nat.findX hm with ⟨m, ⟨hmp, hm₀, a, b, c, d, habcd⟩, hmin⟩
  -- If `m = 1`, then we are done
  rcases (Nat.one_le_iff_ne_zero.2 hm₀.ne').eq_or_gt with rfl | hm₁
  · use a, b, c, d; simpa using habcd
  -- Otherwise, let us find a contradiction
  exfalso
  have : NeZero m := ⟨hm₀.ne'⟩
  by_cases hm : 2 ∣ m
  · -- If `m` is an even number, then `(m / 2) * p` can be represented as a sum of four squares
    rcases hm with ⟨m, rfl⟩
    rw [zero_lt_mul_left two_pos] at hm₀
    have hm₂ : m < 2 * m := by simpa [two_mul]
    apply_fun (Nat.cast : ℕ → ℤ) at habcd
    push_cast [mul_assoc] at habcd
    obtain ⟨_, _, _, _, h⟩ := sum_four_squares_of_two_mul_sum_four_squares habcd
    exact hmin m hm₂ ⟨hm₂.trans hmp, hm₀, _, _, _, _, natAbs_iff.2 h⟩
  · -- For each `x` in `a`, `b`, `c`, `d`, take a number `f x ≡ x [ZMOD m]` with least possible
    -- absolute value
    obtain ⟨f, hf_lt, hf_mod⟩ : ∃ f : ℕ → ℤ, (∀ x, 2 * (f x).natAbs < m) ∧ ∀ x, (f x : ZMod m) = x
    · refine ⟨fun x ↦ (x : ZMod m).valMinAbs, fun x ↦ ?_, fun x ↦ (x : ZMod m).coe_valMinAbs⟩
      exact (mul_le_mul' le_rfl (x : ZMod m).natAbs_valMinAbs_le).trans_lt
        (Nat.mul_div_lt_iff_not_dvd.2 hm)
    -- Since `|f x| ^ 2 = (f x) ^ 2 ≡ x ^ 2 [ZMOD m]`, we have
    -- `m ∣ |f a| ^ 2 + |f b| ^ 2 + |f c| ^ 2 + |f d| ^ 2`
    obtain ⟨r, hr⟩ : m ∣ (f a).natAbs ^ 2 + (f b).natAbs ^ 2 + (f c).natAbs ^ 2 + (f d).natAbs ^ 2
    · simp only [← Int.coe_nat_dvd, ← ZMod.int_cast_zmod_eq_zero_iff_dvd]
      push_cast [hf_mod, sq_abs]
      norm_cast
      simp [habcd]
    -- The quotient `r` is not zero, because otherwise `f a = f b = f c = f d = 0`, hence
    -- `m` divides each `a`, `b`, `c`, `d`, thus `m ∣ p` which is impossible.
    

Proof state:
case neg.intro.intro.intro
p : ℕ
hp : Prime p
this✝ : Fact (Prime p)
natAbs_iff :
  ∀ {a b c d : ℤ} {k : ℕ},
    natAbs a ^ 2 + natAbs b ^ 2 + natAbs c ^ 2 + natAbs d ^ 2 = k ↔ a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = ↑k
hm✝ : ∃ m < p, 0 < m ∧ ∃ a b c d, a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = m * p
m : ℕ
hmin : ∀ m_1 < m, ¬(m_1 < p ∧ 0 < m_1 ∧ ∃ a b c d, a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = m_1 * p)
hmp : m < p
hm₀ : 0 < m
a b c d : ℕ
habcd : a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 = m * p
hm₁ : 1 < m
this : NeZero m
hm : ¬2 ∣ m
f : ℕ → ℤ
hf_lt : ∀ (x : ℕ), 2 * natAbs (f x) < m
hf_mod : ∀ (x : ℕ), ↑(f x) = ↑x
r : ℕ
hr : natAbs (f a) ^ 2 + natAbs (f b) ^ 2 + natAbs (f c) ^ 2 + natAbs (f d) ^ 2 = m * r
⊢ False

Next tactic:
rcases (zero_le r).eq_or_gt with rfl | hr₀