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

Question

question 2: Try to find a positive integer n such that n^2-n+41 is not a prime.

Answer 1

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.

 

e.g.

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

Answer 2

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

Answer 3

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.