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