Thursday, 28 April 2016

number theory - Factorization of polynomial with prime coefficients



I'm interested in the problem linked with this answer.



Let f(x)=an+a1x++an1xn1 be polynomial with distinct ai which are primes.




(Polynomials like that for n=4    f(x)=7+11x+17x2+19x3)




  • Is it for some n possible that xn1 and f(x) have some common divisors?



(Negative answer would mean that it is possible to generate circulant non-singular matrices with any prime numbers)



In other words





  • xn1 has roots which lie (as complex vectors) symmetrically in complex plane on the
    unit circle, can such root be also a root of f(x)=an+a1x++an1xn1 in general case where ai are constrained
    as above?


Answer



This is certainly possible. The easiest way is to use pairs of twin primes and n=4. Such as
f(x)=7+5x+11x2+13x3



where f(1)=0 and x+1 is a common factor of f(x) and x41.






Extending the same idea to third roots of unity. Consider
f(x)=7+5x+17x2+29x3+31x4+19x5.


Because 7+29=5+31=17+19=36 we easily see that the third roots of unity ω=e±2πi/3 are zeros of f(x) as f(ω)=36(1+ω+ω2)=0. Therefore f(x) has a common factor Φ3(x)=x2+x+1 with x31 and therefore also with x61.







For an example of an odd n I found the following. As
53=3+13+37=5+17+31=11+19+23

is the sum of three disjoint triples of primes, we can, as above, show that
f(x)=3+5x+11x2+13x3+17x4+19x5+37x6+31x7+23x8

has the common factor x2+x+1 with x91.


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