= Solution
Let $\mathbf r=\mathbf r_2-\mathbf r_1$ and $\mathbf v=\dot{\mathbf r}$. Relative to the <centre of mass>, the two positions are $\mathbf r_1=-(M_2/M)\mathbf r$ and $\mathbf r_2=(M_1/M)\mathbf r$. Their velocities have the same mass factors. Adding their kinetic energies and <angular momenta> gives, with <reduced mass> $\mu=M_1M_2/M$,
$$
\boxed{E=\frac12\mu v^2-\frac{GM_1M_2}{r}},\qquad
\boxed{\mathbf J=\mu\mathbf r\times\mathbf v=\mu\mathbf h}.
$$
These are the internal orbital quantities, excluding any uniform centre-of-mass motion. <Newton's law of universal gravitation> gives
$$
\ddot{\mathbf r}_1=\frac{GM_2}{r^3}\mathbf r,\qquad
\ddot{\mathbf r}_2=-\frac{GM_1}{r^3}\mathbf r,\qquad
\ddot{\mathbf r}=-\frac{GM}{r^3}\mathbf r.
$$
For constant masses, differentiating $E$ makes the gravitational work cancel the derivative of the <potential energy>, so $\dot E=0$. Also $\dot{\mathbf h}=\mathbf r\times\ddot{\mathbf r}=0$, since the force is central; hence <conservation of angular momentum> gives $\dot{\mathbf J}=0$.
The <eccentricity vector>, the normalized attractive <Laplace-Runge-Lenz vector>, is
$$
GM\mathbf e=\mathbf v\times\mathbf h-GM\widehat{\mathbf r}.
$$
Using $\dot{\widehat{\mathbf r}}=\mathbf v/r-\mathbf r(\mathbf r\mathbin\cdot\mathbf v)/r^3$ and
$$
-\frac{GM}{r^3}\mathbf r\times\mathbf h
=GM\left(\frac{\mathbf v}{r}-\frac{\mathbf r(\mathbf r\mathbin\cdot\mathbf v)}{r^3}\right)
$$
shows that its derivative also vanishes. Adding a relative perturbing acceleration $\mathbf f$ changes these cancellations only by the perturbing terms. The <perturbed Kepler-orbit conservation laws> are therefore
$$
\boxed{\dot E=\mu\mathbf v\mathbin\cdot\mathbf f},\qquad
\boxed{\dot{\mathbf h}=\mathbf r\times\mathbf f},\qquad
\boxed{GM\dot{\mathbf e}=\mathbf f\times\mathbf h+\mathbf v\times(\mathbf r\times\mathbf f)}.
$$
For the distant <Newtonian gravitational potential>, expand each source term about the binary's <centre of mass>:
$$
\frac1{|\mathbf x-\mathbf r_i|}
=\frac1x+\frac{\widehat{\mathbf x}\mathbin\cdot\mathbf r_i}{x^2}
+\frac{3(\widehat{\mathbf x}\mathbin\cdot\mathbf r_i)^2-r_i^2}{2x^3}
+O\left(\frac{a^3}{x^4}\right).
$$
The dipole vanishes because $\sum_iM_i\mathbf r_i=0$. With $\mathbf a=\mathbf r$, the quadratic mass moment is $\sum_iM_i\mathbf r_i\mathbf r_i=\mu\mathbf a\mathbf a$, so the <quadrupole potential of a circular binary> is
$$
\boxed{\phi(\mathbf x)=-\frac{GM}{x}-\frac{G\mu a^2}{2x^3}(3\cos^2\theta-1)+O\left(\frac{GMa^3}{x^4}\right)}.
$$
On the inner binary's fast timescale, the distant companion's position is approximately fixed. Averaging over the inner <circular orbit> in the common plane gives $\langle\cos^2\theta\rangle=1/2$ and hence
$$
\overline\phi(x)=-\frac{GM}{x}-\frac{G\mu a^2}{4x^3}.
$$
This averaged <potential function> is independent of both time and azimuth. Consequently the companion's energy and axial <angular momentum> are conserved in the averaged problem. The exact motion has small fast variations; the statement is a secular approximation for a hierarchical system away from an orbital resonance.
For a circular outer orbit, force balance gives
$$
x\Omega_3^2=\frac{d\overline\phi}{dx}
=\frac{GM}{x^2}+\frac{3G\mu a^2}{4x^4},
$$
so the <circumbinary orbital-period correction> is
$$
\boxed{P_3=2\pi\sqrt{\frac{x^3}{GM}}
\left(1+\frac{3\mu a^2}{4Mx^2}\right)^{-1/2}}
=2\pi\sqrt{\frac{x^3}{GM}}\left[1-\frac{3\mu a^2}{8Mx^2}+O\left(\frac{a^4}{x^4}\right)\right].
$$
<Conservative binary mass transfer> leaves $M$ unchanged, so it does not change the leading monopole force. It changes $\mu$ and $a$, and therefore the quadrupole correction. If the inner orbital <angular momentum> is conserved, $J_{\rm in}=\mu\sqrt{GMa}$ gives $a\propto\mu^{-2}$ and $Q=\mu a^2\propto\mu^{-3}$. The correction is thus largest as the binary becomes strongly unequal in mass and widens, within the hierarchy $a\ll x$.
One must also allow the outer radius to respond. For slow axisymmetric evolution its <specific angular momentum> $h_3$ remains constant, while its energy need not: the averaged potential now depends on time. A nearly circular outer orbit has $h_3^2=GMx+3GQ/(4x)$ and $P_3=2\pi x^2/h_3$. At fixed $M,h_3$, differentiation gives
$$
\frac{dP_3}{P_3}=-\frac{6\,dQ}{4Mx^2-3Q}
\simeq-\frac{3\,dQ}{2Mx^2}.
$$
Thus mass exchange inside the binary can produce a small measurable outer-period change even without total mass loss. Nonconservative mass loss additionally changes the leading monopole term.
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