/-- 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.
    rcases (zero_le r).eq_or_gt with rfl | hr₀
    · replace hr : f a = 0 ∧ f b = 0 ∧ f c = 0 ∧ f d = 0; · simpa [and_assoc] using hr
      obtain ⟨⟨a, rfl⟩, ⟨b, rfl⟩, ⟨c, rfl⟩, ⟨d, rfl⟩⟩ : m ∣ a ∧ m ∣ b ∧ m ∣ c ∧ m ∣ d
      · simp only [← ZMod.nat_cast_zmod_eq_zero_iff_dvd, ← hf_mod, hr, Int.cast_zero, and_self]
      have : m * m ∣ m * p := habcd ▸ ⟨a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2, by ring⟩
      rw [mul_dvd_mul_iff_left hm₀.ne'] at this
      exact (hp.eq_one_or_self_of_dvd _ this).elim hm₁.ne' hmp.ne
    -- Since `2 * |f x| < m` for each `x ∈ {a, b, c, d}`, we have `r < m`
    have hrm : r < m
    · rw [mul_comm] at hr
      apply lt_of_sum_four_squares_eq_mul hr <;> apply hf_lt
    -- Now it suffices to represent `r * p` as a sum of four squares
    -- More precisely, we will represent `(m * r) * (m * p)` as a sum of squares of four numbers,
    -- each of them is divisible by `m`
    rsuffices ⟨w, x, y, z, hw, hx, hy, hz, h⟩ : ∃ w x y z : ℤ, ↑m ∣ w ∧ ↑m ∣ x ∧ ↑m ∣ y ∧ ↑m ∣ z ∧
      w ^ 2 + x ^ 2 + y ^ 2 + z ^ 2 = ↑(m * r) * ↑(m * p)
    · have : (w / m) ^ 2 + (x / m) ^ 2 + (y / m) ^ 2 + (z / m) ^ 2 = ↑(r * p)
      · refine mul_left_cancel₀ (pow_ne_zero 2 (Nat.cast_ne_zero.2 hm₀.ne')) ?_
        conv_rhs => rw [← Nat.cast_pow, ← Nat.cast_mul, sq m, mul_mul_mul_comm, Nat.cast_mul, ← h]
        simp only [mul_add, ← mul_pow, Int.mul_ediv_cancel', *]
      rw [← natAbs_iff] at this
      exact hmin r hrm ⟨hrm.trans hmp, hr₀, _, _, _, _, this⟩
    -- To do the last step, we apply the Euler's four square identity once more
    replace hr : (f b) ^ 2 + (f a) ^ 2 + (f d) ^ 2 + (-f c) ^ 2 = ↑(m * r)
    · rw [← natAbs_iff, natAbs_neg, ← hr]
      ac_rfl
    

Proof state:
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
hr₀ : 0 < r
hrm : r < m
hr : f b ^ 2 + f a ^ 2 + f d ^ 2 + (-f c) ^ 2 = ↑(m * r)
⊢ ∃ w x y z, ↑m ∣ w ∧ ↑m ∣ x ∧ ↑m ∣ y ∧ ↑m ∣ z ∧ w ^ 2 + x ^ 2 + y ^ 2 + z ^ 2 = ↑(m * r) * ↑(m * p)

Next tactic:
have := congr_arg₂ (· * Nat.cast ·) hr habcd