Prove the following theorem: Suppose p is a prime number, r, s are positive integers and x is an
arbitrary integer. Then we have x^r identical to� x^s (mod p) whenever r is identical to� s (mod 11).for x belongs to an integer
To prove the given theorem, we can use the properties of modular arithmetic.
First, let's assume that r is identical to s (mod 11). This means that there exists a positive integer k such that r - s = 11k. Rewrite this equation as r = s + 11k.
Now, let's consider x^r and x^s (mod p).
Using the property of exponentiation, we have x^r ≡ x^(s + 11k) (mod p).
Next, using the property of modular arithmetic, we can rewrite this expression as x^r ≡ (x^s)(x^(11k)) (mod p).
Now we need to show that x^(11k) ≡ 1 (mod p).
To do this, we'll make use of Fermat's Little Theorem which states that if p is a prime number and x is an integer not divisible by p, then x^(p-1) ≡ 1 (mod p).
Since p is a prime number, p - 1 is divisible by 11 (since 11 is a prime factor of 11). Therefore, we can rewrite p - 1 as 11m, where m is a positive integer.
Using Fermat's Little Theorem, we have x^(11m) ≡ 1 (mod p).
Substituting this back into our earlier expression, we have x^r ≡ (x^s)(x^(11k)) ≡ (x^s)(1) ≡ x^s (mod p).
Hence, we have proven that x^r is identical to x^s (mod p) whenever r is identical to s (mod 11).