I came across this problem when I was going through Yahoo answers.

Show that the values of the polynomial have no prime factors of the form for any .

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- December 7th 2009, 12:36 AMroninproInteger-valued polynomial
I came across this problem when I was going through Yahoo answers.

Show that the values of the polynomial have no prime factors of the form for any . - December 7th 2009, 02:01 AMNonCommAlg
- December 7th 2009, 05:34 AMPaulRS
Note that . So if then

Evidently can't divide since then we'd have which would mean that -since p|(1+n+nē) -

Hence we must have , but then - by Fermat's Little Theorem- which is again a contradiction! - March 9th 2010, 08:51 PMkingwinner
- March 10th 2010, 08:19 PMroninpro
The idea in NonCommAlg's post was to look at the equation , where for some integer . Multiplying both sides by 4 gives . Subtracting 3 from both sides gives . We factor the left hand side: .

My original idea was somewhat similar. I first looked at the quadratic equation: . This can be interpreted as a formula modulo by making a few modifications. We should take to mean and to mean the square root modulo . (If you are not comfortable with this, try actually completing the square and solving for , all modulo .) Then, it follows that a quadratic polynomial modulo has roots if and only if the discriminant is a quadratic residue. In the case of the problem, (it is not a coincidence that this is the same as in the above), and we can apply the reciprocity argument as NonCommAlg outlined already.