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

  1. Sep 17, 2006 #1
    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.

  2. jcsd
  3. Sep 17, 2006 #2
    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
  4. Sep 17, 2006 #3
    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?
  5. Sep 17, 2006 #4


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    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, ...
  6. Sep 17, 2006 #5
    Thanks shmoe! i got it, n = 11, u will get 121 which is not prime! wee thanks!
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