Prove or disprove: n^2 + 41n + 41 is a prime number of every n.?

Question

I have been given a Discrete Structure Class assignment to try and prove or disprove: n^2 + 41n + 41 is a prime number of every n.

Now, I tried the first 3 prime numbers, 2, 3, and 5. And yes, it is PROVEN.

However, is there a more professional (math-wise) way of proving it? (maybe by using direct, indirect, contrapositive or etc..methods???)

Hope anyone out there could shed some light on this problem...Thanks in advance!

Answer 1

Sounds like it's saying for ANY integer you put in, it pops out a prime number. Anyhow, the number 41 disproves this theory (it's also prime if you were looking for a prime number). 41^2+41(41)+41=3403, which is divisible by 41 and 83.

Answer 2

n^2 + 41n + 41 is composite for most n, starting with n = 4, 6, 7, 10, 17, 20, 21, 23, 25, 29, 30, 33, 35, 36, 38, 39, 40, 41, ...



Finding n=4 by trial and error seems so easy that I wonder whether another quadratic was intended. Euler famously showed that n^2 + n + 41 is prime for n = 0..39. This could make somebody think it's always prime, but it's clearly divisible by 41 for n=41. In fact, it's composite for n = 40, 41, 44, 49, 56, 65, 76, 81, ...

Answer 3

You can disprove it by using n = 10



n^2 + 41n + 41 =

(10)^2 +41(10) + 41 =

100 + 410 + 41 =

551



551 is NOT prime (it is divisible by 19 and 29).



Ta dah.

Answer 4

n= 41 gives 41^2 + 41 * 41 + 41 = 41 * 83



disproved