Find the smallest prime positive integer in each of the following. 
  
*
  
*Find the smallest positive integer such that $80-n$ and $80+n$ are prime numbers.  
  
*Find the smallest positive prime number such that $2002-n$ and $2002+n$ are prime numbers.
  

I cannot think of any way other than trying the prime numbers one by one, 
like trying from $2, 3, 5, 7,\ldots...$ but it will probably take forever in case the answer is a big number, any clue please? 
Thanks in advance!
 A: For the first.
1) $n$ is not of the form $3k+1$ otherwise $80+n$ is divisible by $3$.
2) $n$ is not of the form $3k+2$ otherwise $80-n$ is divisible by $3$.
3)So n is a multiple of $3$.
4) $n$ can not have factors $2, 5$. In particular $n$ is odd. So n is an odd multiple of $3$ that is not divisible by $5$.
Examining odd multiples (not divisible by $5$ ) of $3$ is easy. $n=9$ is the answer.
A: For question 2, note that the question asks for a prime number n, unlike question 1. The same reasoning applies as in Fermat's answer in terms of mod 3 analysis:
$2002 \equiv 1 \bmod 3$, therefore


*

*for $n \equiv 1 \bmod 3,$ we have $2002-n \equiv 0 \bmod 3$  (only prime if $2002-n=3$)

*for $n \equiv 2 \bmod 3,$ we have $ 2002+n \equiv 0 \bmod 3$  (never prime)

*for $n \equiv 0 \bmod 3,$ we have $ 2002\pm n \equiv 1 \bmod 3$  (only $n=3$ is prime)


We can only possibly have primes for these for the cases where either $n=3$ or $2002-n=3$. In either case, if $2002-3=1999$ were not prime, there would definitely be no solutions - but it is. So we can check just the two cases, $n=3$ and $n=1999$. $2005$ is not a prime, but $4001$ is, giving the only solution of $n=1999$.
