Solutions in the Ring ℤ/105ℤ

The problem asks for the number of solutions to the equation \(x^2 = 1\) in the ring \(\mathbb{Z}/105\mathbb{Z}\). This ring consists of integers modulo 105. The equation \(x^2 = 1\) is equivalent to the congruence \(x^2 \equiv 1 \pmod{105}\).

Prime Factorization of 105

To solve this congruence, we first find the prime factorization of the modulus, 105. \[105 = 3 \times 5 \times 7\] Since 3, 5, and 7 are distinct prime numbers, the ring \(\mathbb{Z}/105\mathbb{Z}\) is isomorphic to the direct product of the rings \(\mathbb{Z}/3\mathbb{Z}\), \(\mathbb{Z}/5\mathbb{Z}\), and \(\mathbb{Z}/7\mathbb{Z}\). This is a result from the Chinese Remainder Theorem. \[\mathbb{Z}/105\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}\]

Solving Congruences Modulo Prime Factors

The congruence \(x^2 \equiv 1 \pmod{105}\) is equivalent to the system of congruences:

• \(x^2 \equiv 1 \pmod{3}\)
• \(x^2 \equiv 1 \pmod{5}\)
• \(x^2 \equiv 1 \pmod{7}\)

We need to find the number of solutions for each of these individual congruences.

Solutions Modulo 3

Consider the congruence \(x^2 \equiv 1 \pmod{3}\). We check the possible values for \(x\) in \(\mathbb{Z}/3\mathbb{Z}\), which are 0, 1, and 2.

• If \(x=0\), \(x^2 = 0^2 = 0 \not\equiv 1 \pmod{3}\).
• If \(x=1\), \(x^2 = 1^2 = 1 \equiv 1 \pmod{3}\). This is a solution.
• If \(x=2\), \(x^2 = 2^2 = 4 \equiv 1 \pmod{3}\). This is a solution.

Alternatively, in \(\mathbb{Z}/3\mathbb{Z}\), we know \(x^2 \equiv 1\) implies \(x^2 - 1 \equiv 0\), which factors as \((x-1)(x+1) \equiv 0 \pmod{3}\). Since 3 is a prime number, \(\mathbb{Z}/3\mathbb{Z}\) is a field, and a product is zero only if one of the factors is zero. So, \(x-1 \equiv 0 \pmod{3}\) or \(x+1 \equiv 0 \pmod{3}\).

• \(x \equiv 1 \pmod{3}\)
• \(x \equiv -1 \equiv 2 \pmod{3}\)

There are 2 solutions modulo 3.

Solutions Modulo 5

Consider the congruence \(x^2 \equiv 1 \pmod{5}\). We check the possible values for \(x\) in \(\mathbb{Z}/5\mathbb{Z}\), which are 0, 1, 2, 3, and 4.

• If \(x=0\), \(x^2 = 0^2 = 0 \not\equiv 1 \pmod{5}\).
• If \(x=1\), \(x^2 = 1^2 = 1 \equiv 1 \pmod{5}\). This is a solution.
• If \(x=2\), \(x^2 = 2^2 = 4 \not\equiv 1 \pmod{5}\).
• If \(x=3\), \(x^2 = 3^2 = 9 \equiv 4 \not\equiv 1 \pmod{5}\).
• If \(x=4\), \(x^2 = 4^2 = 16 \equiv 1 \pmod{5}\). This is a solution.

Alternatively, in \(\mathbb{Z}/5\mathbb{Z}\), \((x-1)(x+1) \equiv 0 \pmod{5}\). Since 5 is prime, \(x-1 \equiv 0 \pmod{5}\) or \(x+1 \equiv 0 \pmod{5}\).

• \(x \equiv 1 \pmod{5}\)
• \(x \equiv -1 \equiv 4 \pmod{5}\)

There are 2 solutions modulo 5.

Solutions Modulo 7

Consider the congruence \(x^2 \equiv 1 \pmod{7}\). We check the possible values for \(x\) in \(\mathbb{Z}/7\mathbb{Z}\), which are 0, 1, 2, 3, 4, 5, and 6.

• If \(x=0\), \(x^2 = 0^2 = 0 \not\equiv 1 \pmod{7}\).
• If \(x=1\), \(x^2 = 1^2 = 1 \equiv 1 \pmod{7}\). This is a solution.
• If \(x=2\), \(x^2 = 2^2 = 4 \not\equiv 1 \pmod{7}\).
• If \(x=3\), \(x^2 = 3^2 = 9 \equiv 2 \not\equiv 1 \pmod{7}\).
• If \(x=4\), \(x^2 = 4^2 = 16 \equiv 2 \not\equiv 1 \pmod{7}\).
• If \(x=5\), \(x^2 = 5^2 = 25 \equiv 4 \not\equiv 1 \pmod{7}\).
• If \(x=6\), \(x^2 = 6^2 = 36 \equiv 1 \pmod{7}\). This is a solution.

Alternatively, in \(\mathbb{Z}/7\mathbb{Z}\), \((x-1)(x+1) \equiv 0 \pmod{7}\). Since 7 is prime, \(x-1 \equiv 0 \pmod{7}\) or \(x+1 \equiv 0 \pmod{7}\).

• \(x \equiv 1 \pmod{7}\)
• \(x \equiv -1 \equiv 6 \pmod{7}\)

There are 2 solutions modulo 7.

Total Number of Solutions

According to the Chinese Remainder Theorem, for each combination of solutions from the individual congruences modulo the prime factors, there is a unique solution modulo the product of the moduli (which is 105).

The number of solutions modulo 105 is the product of the number of solutions modulo 3, modulo 5, and modulo 7.

Number of solutions = (Solutions mod 3) \(\times\) (Solutions mod 5) \(\times\) (Solutions mod 7)

Number of solutions = \(2 \times 2 \times 2 = 8\).

Thus, there are 8 solutions to the equation \(x^2 = 1\) in the ring \(\mathbb{Z}/105\mathbb{Z}\).