Hint :
If satisfies then,
For , that is, .
For , that is, .
...
are roots of
Fernando Revilla
Hint :
If satisfies then,
For , that is, .
For , that is, .
...
are roots of
Fernando Revilla
Didn't we just recently have this?
Here's how I would do it (rather different from the way FernandoRevilla did it). Taking x= 5, 5p(4)= 0 so that p(4)= 0. Since p is a polynomial, that means that p has a factor of x- 4: we can write p(x)= (x- 4)q(x) for some polynomial q.
Putting that into the given equation, xp(x- 1)= x((x- 1)- 4)q(x-1)= x(x- 5)q(x-1)= (x- 5)(x- 4)q(x). Since this is true for all x, we can cancel the "x- 5" terms an see that q satisfies
xq(x- 1)= (x- 4)q(x).
Talking x= 4, we have 4q(3)= 0 so that q(3)= 0 and we can write q(x)= (x- 3)r(x) for some polynomial r.
Now xq(x-1)= x(x-1- 3)r(x-1)= (x- 4)(x- 3)r(x). Since this is true for all x, we cancel the "x-4" terms and have
xr(x-1)= (x- 3)r(x). Taking x= 3, we have 3r(2)= 0 so that r(x)= (x- 2)s(x) for some polynomial s.
Get the point? Keep doing that until it terminates (it will).
We have proved that are roots of so, all the polynomials satisfying
have the form:
On the other hand, implies:
that is, and necessarily constant ( Why ?).
So, all polynomials satisfying are:
and the condition implies .
Fernando Revilla