= Solution
Writing the <Bogoliubov transformation> componentwise, together with its <Hermitian adjoint>, gives
$$
a=uc+vd^\dagger,\qquad b=-vc^\dagger+ud,\qquad
a^\dagger=uc^\dagger+vd,\qquad b^\dagger=-vc+ud^\dagger.
$$
The full <canonical anticommutation relations> for $c,d$ imply
$$
\{a,a^\dagger\}=u^2\{c,c^\dagger\}+v^2\{d^\dagger,d\}=u^2+v^2,
\qquad \{b,b^\dagger\}=u^2+v^2.
$$
Equal-type <anticommutators> vanish because $c^2=d^2=0$ and $\{c,d^\dagger\}=\{c^\dagger,d\}=0$:
$$
\{a,a\}=\{b,b\}=0.
$$
For the mixed <anticommutators>, the only potentially nonzero terms cancel:
$$
\{a,b\}=-uv\{c,c^\dagger\}+uv\{d^\dagger,d\}
=-uv+uv=0,
$$
while $\{a,b^\dagger\}=0$ follows from $\{c,c\}=\{d^\dagger,d^\dagger\}=\{c,d^\dagger\}=0$. Taking the <Hermitian adjoint> supplies the remaining relations. Thus the necessary and sufficient condition is
$$
\boxed{u^2+v^2=1.}
$$
Equivalently, the <matrix> $R=\begin{pmatrix}u&v\\-v&u\end{pmatrix}$ is an <orthogonal matrix>, $R^{\mathsf T}R=I$, giving a <canonical rotation of two fermionic modes>. Its inverse is therefore
$$
\boxed{c=ua-vb^\dagger,\qquad d=va^\dagger+ub.}
$$
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