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For what value of n are \[{2^n} - \;1\] and \[{2^n} + \;1\] prime?
A) 7
B) 5
C) 2
D) 1

Answer
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Hint: A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. Since values of n given are smaller hit and trial is the best suited method for the question. A prime number is divisible by only 1 and the number itself.
Eg- 2,3,5,7,11,13,17,19,etc.

Complete step-by-step answer:
We are given various values for n and we need to find out for what value \[{2^n} - \;1\] and \[{2^n} + \;1\] are both Prime numbers.
 Since, a prime number is not divisible by any number except 1 and the number itself.
For example: 13 is a prime number because -
13 when divided by 1 = $\dfrac{{13}}{1} = 13$ and when divided by 13 we get $\dfrac{{13}}{{13}} = 1$ gives a whole number. There are no numbers similar to 1 and 13 which can divide 13 and result in a natural number.
 Now, to prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can't be a prime number. If you don't get a whole number, next try dividing it by prime numbers: 3, 5, 7, and 11 (9 is divisible by 3) and so on, always dividing by a prime number.
Now, to solve the question, hit and trial is the best method to approach.
For 7 option (a)
$\begin{gathered}
  {2^7} - 1 = 127(prime) \\
  {2^7} + 1 = 129(non - prime) \\
\end{gathered} $
Since, 129 (divisible by 3) is non-prime 7 is incorrect.
For 5 option (b)
$\begin{gathered}
  {2^5} - 1 = 31(prime) \\
  {2^5} + 1 = 33(non - prime) \\
\end{gathered} $
Since 33 (divisible by 11 and 3) is non-prime, 5 is incorrect.
For 1 option (d)
$\begin{gathered}
  {2^1} - 1 = 1(non - prime) \\
  {2^1} + 1 = 3(prime) \\
\end{gathered} $
Since, 1 is non-prime 5 is incorrect.
For 2 option (c)
$\begin{gathered}
  {2^2} - 1 = 3(prime) \\
  {2^2} + 1 = 5(prime) \\
\end{gathered} $

Since, both the numbers are prime, option (c) is correct.

Note: If for some positive integer n, \[{2^n} - \;1\] is prime, then so is n. In the above question numerical values of n are smaller can be solved by hit and trial. There can be cases when these values become tedious. In such situations using this theorem question can be simplified.