Mysterious N

Speak of $n-1$ ,remember the parity change & rules of prime number

In $n-a$ the parity will change if $a$ odd and stay if $a$ even.Since $1$ is odd,so the parity will change.

The rules of prime numbers is they only divisible by 1 & itself.Since even numbers also divisible by 2,so not even numbers but only 2 satisfies because divisible by 2 is divisible itself.So the answer is $2+1=\boxed{\large{3}}$

One of the two, either n or (n-1), must be even. Only one even is prime, 2. If n is 3, a prime number which would return 1 when plugged into p, then (n-1) would be 2, another number to output 1, which means that it works. No other even is prime and n = 2 gives (n-1) = 1 (which is neither prime nor composite thus returning 0) so $\boxed{3}$ is the only thing that works

We know that 3 works. We consider some other solution $z$ . We know that $z \neq 3$ , and $z$ is prime. Since 2 is the only even prime, and $z$ is not 2 (1( z - 1) isn't prime) we know that $z$ is odd. Therefore, $z$ can be expressed as $z = 2x+1$ for some x. That means that $z-1 = 2x$ , which means that $z - 1$ isn't prime, so the conditions are not met.