= Solution
A $j=0$ state exists only when $j_1=j_2=:j$. Since its total magnetic quantum number is zero, every uncoupled basis term must have $m_1+m_2=0$, so
$$
|0,0\rangle=\sum_{m=-j}^j\alpha_m
|j,m\rangle_1|j,-m\rangle_2.
$$
Apply the total raising operator $J_+=J_{1+}+J_{2+}$. The coefficient of
$|j,m+1\rangle_1|j,-m\rangle_2$ is
$$
\sqrt{(j-m)(j+m+1)}\,(\alpha_m+\alpha_{m+1}).
$$
The <angular momentum singlet state> is annihilated by $J_+$, hence
$$
\alpha_{m+1}=-\alpha_m,
\qquad -j\leq m<j.
$$
All coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to $(2j+1)^{-1/2}$, and an arbitrary overall phase may be chosen so that
$$
\boxed{
\alpha_m=\frac{(-1)^{j-m}}{\sqrt{2j+1}}
}.
$$
Thus
$$
\boxed{|0,0\rangle=\frac1{\sqrt{2j+1}}
\sum_{m=-j}^j(-1)^{j-m}|j,m\rangle_1|j,-m\rangle_2.}
$$
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