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Standard VI

Mathematics

Common Factors

Let n be a ...

Question

Let $n$ be a positive integer such that $2^{n} 1$ is a prime number. Prove that n is a prime number.

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Solution

If n is not a prime number, then n=ab ,
For some positive integers a, b>1, we get
$2^{n} - 1 = 2^{a b} - 1 = (2^{a} - 1) [(2^{a})^{b - 1} (2^{a})^{b - 2} . . . . . a^{1} + 1]]$
So, n is composite and $2^{n} - 1$ is also composite which is a contradiction.
Thus, n is prime

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Let n be a positive integer such that 2n1 is a prime number. Prove that n is a prime number.

If n is not a prime number, then n=ab , For some positive integers a, b>1, we get2n−1=2ab−1=(2a−1)[(2a)b−1(2a)b−2.....a1+1]]So, n is composite and 2n−1 is also composite which is a contradiction.Thus, n is prime

Let $n$ be a positive integer such that $2^{n} 1$ is a prime number. Prove that n is a prime number.

If n is not a prime number, then n=ab ,
For some positive integers a, b>1, we get
$2^{n} - 1 = 2^{a b} - 1 = (2^{a} - 1) [(2^{a})^{b - 1} (2^{a})^{b - 2} . . . . . a^{1} + 1]]$
So, n is composite and $2^{n} - 1$ is also composite which is a contradiction.
Thus, n is prime