Proving/disproving n^2-n+11 is prime, i think i got it

[mr_coffee]

Hello everyone!

I think i got this but I'm not sure if I'm allowed to do this. The question is:

For all integers n, n^2-n+11 is a prime number. Well if that was a prime number it should be true that n^2-n+11 = (r)(s) then r = 1 or s = 1. But if you equate n^2-n+11 = 1, you get a false statement. n^2-n + 12 = 0, and if u plugged say 0 in for n, u get 12 = 0, 12 is not prime...but 12 = 0, doesn't make sense. Am i on the right track or totally doing the wrong test? I'm confused if I'm suppose to set n^2-n+11 to somthing, it won't facotr unless i do a quadtratic but I'm not sure what that would even show me.

Thanks!

 

[vsage]

you don't need to solve a polynomial. Giving any counterexample is enough to prove the statement false. This example is very specific, but it's easy to generalize this to n^2 - n + p, where p is any integer

 

[mr_coffee]

I can't seem to find a number that will make it not prime! I've tried a bunch of random ones but they are turn out to be prime, can anyone clarify that is infact false and can be proved with a counter example?

 

[shmoe]

Can you see a choice of n that will let you factor something out of n^2-n+11? If not, be systematic n=1, 2, 3, 4, ...

 

[mr_coffee]

Thanks shmoe! i got it, n = 11, u will get 121 which is not prime! wee thanks!