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prove that 3 is the only prime which can be written in the form n^2-1 for some integer n.

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3 Answers

n2-1 = (n - 1)(n+1)
A number n2-1 will have the factors n-1, n+1.
If a number has a factor other than 1 and itself, it is not prime.
In the case of 3. (n=2) n-1 and n+1 are 1 and 3. (1 and itself)
If n - 1 is NOT 1. The factor pair (n-1)(n+1) will NOT be 1 and itself.
Meaning, it will have some other integers as factors.
Hence, if n-1 is NOT 1, n2-1 is NOT prime.
n = 5
(n-1) is not 1
n2-1 has integer factors n-1,n+1...4,6
NOT prime
If we only need to find 1 answer for Quest 2. It's fairly simple...
412-41+41  = 412


n2-1=(n-1)(n+1). Since it is the product of two factors, it is not a prime unless either n-1 or n+1 is 1. Hence n=2 or n=0. Since for n=0 n2-1=-1, n=2 is the only possibility, in this case n2-1=22-1=3.
n2-n+41 is not a prime when n2-n is a multiple of 41. That is n2-n=41k, where k is positive integer. Now we have to solve n2-n-41k=0 for n and see which k gives positive integer solution.
n(n-1)=41k; Since 41 is a prime, k=40 or k=42; So n=41 or n=42; In the first case, n2-n+41=412-41+41=412. In the second case, 422-42+41=1763=41*43
N^2-1 can be rewritten as (n-1)(n+1). If n=0, we get -1, which is not prime, nor is 0 (when n=1). When n=2, we get three which is clearly prime. For n>2, however, we get factors greater than one, making (n+1)(n-1) not prime. 
To find a number such that n^2-n+41 is not prime, we take a cue from the previous point that we need to find n such that it can rewritten as some x^2-1. In this case n=42, making the expression (42+1)(42-1)=43*41
Checking out work:42^42-42+41=41*42+41=43*41, not a prime, but a product of two primes.