Proof
We prove the statement about the existence by induction over the
degree
of
. If the degree of
is larger than the degree of
, then
and
is a solution.
Suppose that
.
By the remark just made also
holds, so
is a constant polynomial, and therefore (since
and
is a field)
and
is a solution.
So suppose now that
and that the statement for smaller degrees is already proven. We write
and
with
. Then setting
we have the relation
![{\displaystyle {}{\begin{aligned}P'&:=P-TH\\&=0X^{n}+{\left(a_{n-1}-{\frac {a_{n}}{b_{k}}}b_{k-1}\right)}X^{n-1}+\cdots +{\left(a_{n-k}-{\frac {a_{n}}{b_{k}}}b_{0}\right)}X^{n-k}+a_{n-k-1}X^{n-k-1}+\cdots +a_{0}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/295d6edbb2a66a4b486b5373ee6537893a27ac5c)
The degree of this polynomial
is smaller than
and we can apply the induction hypothesis to it. That means there exist
and
such that
-
From this we get altogether
-
![{\displaystyle {}P=P'+TH=TQ'+TH+R'=T(Q'+H)+R'\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d2ab25df331ffc1420888bfc7c650a5a9a910e6)
so that
and
is a solution.
To prove uniqueness, let
,
both fulfilling the stated conditions. Then
.
Since the degree of the difference
is smaller than
, this implies
and so
.