Today, we have another bit of number theory: what are the positive integers *n* such that *n*^{3}+1 is prime?

Well, we recall that *a*^{3}+*b*^{3}=(*a*+*b*)(*a*^{2}–*a**b*+*b*^{2}). Thus, *n*^{3}+1=(*n*+1)(*n*^{2}–*n*+1).

Now, for *n*^{3}+1 to be prime, one of those factors must be equal to one. For positive n, *n*+1>1, and so we have *n*^{2}–*n*+1=1, which means

*n*^{2}–*n*=0

*n*(*n*-1)=0

which, for positive *n*, gives only *n*=1, for *n*^{3}+1=1^{3}+1=2 prime. All other positive *n* give composite numbers.

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Tags: Factoring, Math, Monday Math, Number Theory

This entry was posted on August 25, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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