Bunuel wrote:
Is the positive integer x even?
(1) (x - 1) is a prime number
(2) (x^2 - 1) is a prime number
From the question stem, we get that:
\(x\) is an integer and \(x>0\)
Now lets consider the statements:
Statement 1
\((x - 1)\) is a prime number
This gives us 2 scenarios:
either \((x - 1)\) is an even prime number (i.e. 2) or \((x - 1)\) is an odd primer number (i.e. any prime numbers except 2)
If \((x - 1)\) is an even prime number, then \(x = 3\), an odd integer
If \((x - 1)\) is an odd prime number, then \(x\) has to be an even integer
Therefore, we do not get a definitive yes/no from this statement. This statement is insufficient.
Hence, we can reject options A and D.
Statement 2
\((x^2 - 1)\) is a prime number.
This also gives us 2 scenarios:
either \((x^2 - 1)\) is an even prime number (i.e. 2) or \((x^2 - 1)\) is an odd primer number (i.e. any prime numbers except 2)
If \((x^2 - 1) = 2\); \(x^2 = 3\). This is not possible as we know that \(x\) is an integer.
Hence, the only possible scenario is that \((x^2 - 1)\) is an odd primer number.
This implies \(x^2\) is an even number and accordingly, \(x\) is an even number.
This statement is sufficient. We can reject options C and E.
The correct answer is
B.