Determine whether or not N is a composite number, where

NOTE:
A prime number (or a prime) is a natural number that has exactly two (distinct) natural number divisors, which are 1 and the prime number itself. A composite number is a positive integer which has a positive divisor other than one or itself. By definition, every integer greater than one is either a prime number or a composite number. The numbers 0 and 1 are considered to be neither prime nor composite.

(In reply to A first stab at the solution by e.g.)

I supposed there had to be some factoring of x^23-x^21+x^12+x^10-1. I decided to go the easy way, and look for two factors like a.x^p+b.x^q+1 and c.x^r+d.x^s-1. (At least, the -1 would be produced.)

Multiplying suggests that c=1/1 and r=23-p, so now we are looking at a.x^p+b.x^q+1 and (1/a).x^(23-p)+b.x^s-1. Doing the multiplication produces SIX terms, that reduce to five if 23-p+q=p. In order to get the other exponents to match, p+s=21 and q+s=10, which finally produces p=12, q=1, and s=9.

Equating terms, finally gets to a=-b=-2 and c=-d=-1/2. So, the original polynomial is the same as (-2.x^12+2x+1) multiplied by (-1/2)x^11+(1/2).x^9-1, and for x=26, both terms are integer and way greater than 1, so N is composite!

Posted by Federico Kereki on 2006-04-23 19:10:55