= Solution
Let $m,q,n_s$ denote pair mass, charge and number density, respectively. A uniform-phase condensate in the stated gauge has zero <canonical momentum> $m\mathbf v_s+q\mathbf A=0$. Therefore
$$
\mathbf j_s=n_sq\mathbf v_s=-\frac{n_sq^2}{m}\mathbf A,\qquad \nabla\times\mathbf j_s=-\frac{n_sq^2}{m}\mathbf B.
$$
For static fields, $\nabla\times\mathbf B=\mu_0\mathbf j_s$ and $\nabla\cdot\mathbf B=0$. Taking a curl and using the vector identity for a double curl gives
$$
-\nabla^2\mathbf B=-\frac{\mu_0n_sq^2}{m}\mathbf B,\qquad \boxed{\nabla^2\mathbf B=\frac{\mathbf B}{\lambda^2},\quad\lambda=\sqrt{\frac m{\mu_0n_sq^2}}}.
$$
For electronic pairs, $m\simeq2m_e$ and $q=-2e$, while $n_s$ is the pair density, not twice that density. In a general gauge the <canonical momentum> is $\hbar\nabla\theta$; taking the curl away from vortices gives the same London equation.
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