Integers of the form (1+2x+4x^3)/(x+n)

[jb at brennen.net]

Zak wrote:
> No solutions for n>10?

In general, in PARI/GP, you can do this:

> (4*x^3+2*x+1)%(x+10)
%1 = -4019
> divisors(4019)
%2 = [1, 4019]

The quantity (x+10) must divide into 4019, basically,
and since 4019 is prime, that leaves 4 possibilities
for x:  -4029, -11, -9, and 4009.

You should probably check that all 4 of these
x values lead to different integers, and they do,
so there are 4 solutions for n=10.

   Jack