I'm interested in the problem linked with this answer.
Let f(x)=an+a1x+⋯+an−1xn−1 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 xn−1 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
- xn−1 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+⋯+an−1xn−1 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 x4−1.
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 x3−1 and therefore also with x6−1.
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 x9−1.
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