= Solution
{c}
Clearing the denominators in the <rational interpolation> conditions gives
$$
\boxed{\sum_{j=0}^m a_jx_i^j+
f_i\sum_{j=0}^{\ell-1}b_jx_i^j=f_ix_i^\ell,
\qquad i=0,\ldots,n.}
$$
These are $n+1$ linear equations for $(m+1)+\ell=n+1$ unknowns. The $i$th row of the coefficient <matrix>, in the order $a_0,\ldots,a_m,b_0,\ldots,b_{\ell-1}$, is
$$
\left(1,x_i,\ldots,x_i^m,\ f_i,f_ix_i,\ldots,f_ix_i^{\ell-1}\right).
$$
This <rational interpolation linearization> is equivalent to the original fraction conditions only if $r(x_i)\ne0$ at every <interpolation node>. If $r(x_i)=0$, the cleared equation merely forces $q(x_i)=0$, leaving the fraction undefined.
For $f_i=x_i^{m+1}$, the coefficient row becomes
$$
(1,x_i,\ldots,x_i^m,x_i^{m+1},\ldots,x_i^{m+\ell})
=(1,x_i,\ldots,x_i^n).
$$
This is the <Vandermonde matrix>, whose <determinant> is $\prod_{0\leq i<j\leq n}(x_j-x_i)\ne0$. Equivalently, a vector in its <null space> would define a degree-at-most-$n$ <polynomial> vanishing at all $n+1$ distinct <interpolation nodes>, which is zero by the preceding uniqueness proof. Thus the cleared <linear system> has exactly one solution.
There is also a constructive description of that solution. Define the <monic polynomial>
$$
W(x)=\prod_{i=0}^n(x-x_i).
$$
The <polynomial> $x^{m+1}r(x)-q(x)$ is monic of degree $n+1$ and vanishes at every <interpolation node>. It must therefore equal $W(x)$. By <polynomial division>, write uniquely
$$
W(x)=x^{m+1}R(x)+S(x),\qquad\deg S\leq m.
$$
The quotient $R$ is monic of degree $\ell$, and
$$
\boxed{r(x)=R(x),\qquad q(x)=-S(x)}
$$
is the unique coefficient solution. This does not assert that $R$ is nonzero at every <interpolation node>.
For a counterexample with $m=0$, $\ell=1$, choose $x_0=0,x_1=1$ and the specified data $f_i=x_i$, so $f_0=0,f_1=1$. The cleared equations reduce to
$$
a_0=0,\qquad a_0+b_0=1,
$$
giving the unique solution $q(x)=0$, $r(x)=x-1$. But at $x_1=1$ the quotient is $0/0$. Indeed, any genuine <rational interpolant> of this prescribed form would need $r(0)\ne0$; then $q(0)/r(0)=0$ forces the constant numerator to be zero. At a nonzero denominator its value at $1$ would also be zero, never one. Cancelling the common factor in the cleared candidate would merely give the identically zero <rational function>, which still fails the data. Hence
$$
\boxed{(x_0,f_0)=(0,0),\ (x_1,f_1)=(1,1):
\quad\text{unique cleared solution, but no rational interpolant}.}
$$
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