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

`(x^80+x^48+x^24+x^8+1)/(x^2-1)`

If we try long division on this we soon see a pattern:

x^78+x^76+x^74+x^72+x^70+x^68+...

x^2-1|x^80

-(x^80-x^78)

x^78

-(x^78-x^76)

x^76

-(x^76-x^74)

x^74 ...

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+

-(x^50-x^48)

2x^48

-(2x^48-2x^46)

2x^46 ...

This pattern continues (12 terms with coefficient 2) until we reach x^24:

...+ 2x^24+3x^22+3x^20+3x^18+...

x^2-1| ... x^24 + ...

-(2x^26-2x^24)

3x^24

-(3x^24-3x^22)

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

-(3x^10-3x^8)

4x^8

-(4x^8-4x^6)

4x^6

-(4x^6-4x^4)

4x^4

-(4x^2-4x^2)

4x^2 + 1

-(4x^2 -4)

5

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.

`5/(x^2-1)=(5x)/(x^3-x)`