I learned electrodynamics. According to the vector potential determination, $$ \mathbf B = [\nabla \times \mathbf A ], $$ Coulomb gauge, $$ \nabla \mathbf A = 0, $$ and one of Maxwell's equations, $$ [\nabla \times \mathbf B ] = \frac{1}{c}4\pi \mathbf j, $$
I can assume, that
$$ [\nabla \times \mathbf B ] = \nabla (\nabla \mathbf A) - \Delta \mathbf A = -\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j. $$ How to prove that the one of the solutions of this equation is solution like newtonian potential, $$ \mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|}? $$