Question

Suppose n is a large odd number. You calculate 2 ⁽ⁿ−¹⁾/²≡k(mod n), where k is an integer with k≠±1(modn). Show that n must be composite.

Answer 1

Using Fermat's Little Theorem and the base 2 Fermat test, we can deduce that if 2^(n-1)/2 ≃ k (mod n) for k ≠ ±1 (mod n), then n must be composite since it fails to meet the theorem's condition for a prime number.

Step-by-step explanation:

The given scenario involves testing whether a number is prime through a Fermat's Little Theorem related process known as the base 2 Fermat test. According to Fermat's theorem, if n is a prime number and a is any positive integer less than n, then a raised to the power of n-1 will be congruent to 1 modulo n, or mathematically a^(n-1) ≃ 1 (mod n). Specifically, when a is 2, for a large odd prime n, 2^(n-1)/2 should equal 1 when taken modulo n.

However, in your case, you have calculated 2^(n-1)/2 ≃ k (mod n), where k is not equal to ±1 (mod n). This result already suggests that n cannot be prime because it violates Fermat's Little Theorem for the base 2. Therefore, the fact that you are not getting ±1 as the result implies that n must be composite.