The remainder when `x^81+x^49+x^25+x^9+x` is divided by `x^3-x` is 5x.
We can rewrite as `(x(x^80+x^48+x^24+x^8+1))/(x(x^2-1))` and as long as `x ne 0` we get
If we try long division on this we soon see a pattern:
This pattern continues (16 terms with coefficient 1) until we reach x^48:
... + x^48+2x^46+2x^44+2x^42+...
x^2-1| ... x^48+
This pattern continues (12 terms with coefficient 2) until we reach x^24:
x^2-1| ... x^24 + ...
3x^22 + ...
This pattern continues (8 terms with coefficient 3) until we reach x^8:
... + 3x^8+4x^6+4x^4+4x^2+5
x^2-1| ... + x^8 + + 1
4x^2 + 1
So we have `x^81+x^49+x^25+x^9+x)=(Q)(x^3-x)+5x` (Note that we divided by x^2-1 with a remainder of 5, so the remainder is 5x.)
Q is a polynomial where the powers of x start at 78 and descend by twos, and the coefficients are 16 1's, 12 2's, 8 3's, 4 4's, and a remainder of 5x.