For a positive integer n, by using Chinese Remainder Theorem, the number of solutions of the congruence x2 ≡ 1 (mod n), when e ≥ 3 is:

For a positive integer n, by using Chinese Remainder Theorem, the number of solutions of the congruence x2 ≡ 1 (mod n), when e ≥ 3 is: Correct Answer 2^{k + 2}

Concept:

By the Chinese Remainder theorem, the number of solution of x2 ≡ 1 (mod p^k) is 2^k+1.

Calculation:

As we have given x2 ≡ 1 (mod n)

The above congruence means x² - 1 is divisible by n.

⇒ n | x² - 1

or n | (x - 1) (x + 1)

Case I, when n = 1;

When n = 1 then there is only two solutions of x2 ≡ 1 (mod n).

Then,

⇒ 1 | (x - 1)(x + 1) 

so, there are two solutions to the above.

Case II, when n > 1;

Let, n = p^k as any number can be written in the form of the product of prime numbers.

Then above congruence can be written as

⇒ x2 ≡ 1 (mod p^k)

The above-given congruence has a 2^k+1 number of solution, by the Chinese Remainder theorem.

Therefore, there is a total  2k+2 solution of the above congruence relation, 2 solutions for n = 1 and  2k+1 number of solution for n > 1.

Hence, the number of solution of the congruence is 2k+2.