Crt counting

• A.Wrong

  False — x = 1 is always a solution

  so the count is never 0.

• B.WrongBase field or ring

  False.

  ±1 are the obvious two

  but they are not the only ones once the modulus is composite.

  SymPy exhibits others directly: $29^{2} \equiv 1$ and $34^{2} \equiv 1 (mod 105)$.

  Expecting exactly two square roots of 1 is the habit carried over from fields, where $x^{2} - 1 = (x-1)(x+1)$ has at most two roots — an argument that needs the absence of zero divisors.

• C.WrongExecution slip

  False — 4 would be the count for a product of two odd primes.

  105 has three distinct prime factors (SymPy: 3 · 5 · 7)

  so the count is $2^{3}$, not $2^{2}$.

• D.Correct

  True.

  By the Chinese remainder theorem $\mathbb{Z}/105 \cong \mathbb{Z}/3 \times \mathbb{Z}/5 \times \mathbb{Z}/7 ($SymPy confirms the factorisation), and $x^{2} = 1$ becomes three independent equations, one per component.

  Each is over a field of odd characteristic, where $x^{2} = 1$ has exactly the two roots ±1.

  So the total is 2 × 2 × 2 = 8 (verified).

  The trigger: for a modular equation, split by CRT and solve prime by prime, then multiply the counts.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.