x=n^2-1 x is a prime number and n is a natural number find x^n
Answers
Answer:
1. Homework Statement
I was able to prove both of these statements after getting some help from another website, but I am trying to find another way to prove them. Can you guys check my work and help me find another way to prove these, if possible? Thanks.
Part A: Show that if 2^n - 1 is prime, then n must be prime.
Part B: Show that if 2^n + 1 is prime, where n
≥
≥
1, then n must be of the form 2^k for some positive integer k.
2. Homework Equations
(x^k) - 1 = (x - 1)*(x^(k-1) + x^(k-2) + ... + x + 1)
3. The Attempt at a Solution
Part A:
Write the contrapositive,
n is not prime (a.k.a. n is composite) ==> 2^n - 1 is composite
Assume n is composite. Let n = p*q, where neither p nor q are 1.
Then,
2^n - 1 = (2^p)^q - 1 = (2^p - 1)*((2^p)^(q-1) + (2^p)^(q-2) + ... + (2^p) + 1)
Note that 2^p - 1 > 1. Also, ((2^p)^(q-1) + (2^p)^(q-2) + ... + (2^p) + 1) > 1. So we have factored 2^n - 1, thus it is not prime. We have proved the contrapositive, so the original statement is true.
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Part B:
Note that if n is of the form 2^k, then n's prime factorization is only composed of 2's. Thus, the contrapositive of the original statement is as follows:
n = b*(2^k), where b is a positive odd number ==> 2^n + 1 is composite.
Let n = b*(2^k). Then,
2^n + 1 = 2^(b*(2^k)) + 1 = ((2^(2^k))^b + 1 = (2^(2^k) + 1)*{[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1}
Observe that [2^(2^k) + 1)] > 1 and {[(2^(2^k)]^(k-1) + [2^(2^k)]^(k-2) + ... + [2^(2^k)] + 1} > 1. We have factored 2^n + 1, so it is composite. This proves the contrapositive of the original statement, so the original statement is true.
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Answer:
When n is equal to 1 then then x also will be 1
When n is 2 then x is 3.
And so on...
u will get xn