^{2}-1 = (n - 1)(n+1)

^{2}-1 will have the factors n-1, n+1.

^{2}-1 is NOT prime.

^{2}-1 has integer factors n-1,n+1...4,6

^{2}-41+41 = 41

^{2}

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

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Eric Y. | SAT PrepSAT Prep

n^{2}-1 = (n - 1)(n+1)

A number n^{2}-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, n^{2}-1 is NOT prime.

e.g.

n = 5

(n-1) is not 1

n^{2}-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...

41^{2}-41+41 = 41^{2}

n^{2}-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 n^{2}-1=-1, n=2 is the only possibility, in this case n^{2}-1=2^{2}-1=3.

n^{2}-n+41 is not a prime when n^{2}-n is a multiple of 41. That is n^{2}-n=41k, where k is positive integer. Now we have to solve n^{2}-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, n^{2}-n+41=41^{2}-41+41=41^{2}. In the second case, 42^{2}-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.

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