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>		 Find the number of polynomials of degree 5 with distinct coefficients from the set {1,2,3,4,5,6,7,8} that are divisible by x²-x+1.
Let ax^5+bx^4+cx^3+dx+2+ex+f be a polynomial.  Using long division, dividing it by x²-x+1, we get a remainder of

(-a-b+d+e)x + (-c-d+f+a)

The remainder is zero, so

-a-b+d+e=0 and -c-d+f+a=0

Rearranging,

a+b=d+e and c+d=f+a and combining these, b+c=e+f as well.

It has been pointed out that this last criterion is redundant, but I mention it because it completes a symmetry.

These are some wonderfully symmetrical criteria. If you arrange the numbers a through f on the vertices of a hexagon, and label each side with the sum of the vertices at its endpoints, then parallel sides have equal labels.

That means the 12 rotations and reflections of the hexagon will sort the solutions in classes of 12.

Now, running low on ideas, I listed all 20160 permutations of the 8 elements of the set, taken 6 at a time, and found that there are 300 that meet the criteria, so if I arrange the coefficients of each of these 300 polynomials on the vertices of 300 hexagons, I should find 25 groups of 12 rotations and reflections.

I decided to list each of the 25 groups by naming the lowest-numbered vertex followed by its lower-numbered neighbor, to see if I could spot a pattern that would give me a clue about counting them.

The groups are:

1, 4, 5, 2, 3, 6
1, 4, 5, 3, 2, 7
1, 4, 6, 2, 3, 7
1, 4, 6, 3, 2, 8
1, 4, 7, 2, 3, 8
1, 5, 3, 4, 2, 6
1, 5, 6, 2, 4, 7
1, 5, 7, 2, 4, 8
1, 6, 3, 5, 2, 7
1, 6, 4, 5, 2, 8
1, 6, 5, 3, 4, 7
1, 6, 5, 4, 3, 8
1, 6, 7, 2, 5, 8
1, 7, 3, 6, 2, 8
1, 7, 4, 5, 3, 8
1, 7, 6, 3, 5, 8
2, 5, 6, 3, 4, 7
2, 5, 6, 4, 3, 8
2, 5, 7, 3, 4, 8
2, 6, 4, 5, 3, 7
2, 6, 7, 3, 5, 8
2, 7, 4, 6, 3, 8
2, 7, 6, 4, 5, 8
3, 6, 7, 4, 5, 8
3, 7, 5, 6, 4, 8

But I'm still at a loss to describe how I would count these polynomials and get a total of 300, or how I would count the unique (up to rotation and reflection) hexagons and get a total of 25.

Still stuck . . . . . . . .