= Solution
In elastic diagonal <factorized scattering>, the <Faddeev-Zamolodchikov algebra> describes an exchange of two particle operators. Exchanging the pair twice must restore the original state. This is analytic <unitarity>:
$$
S_{AA}(\theta)S_{AA}(-\theta)=1,\qquad S_{A\bar A}(\theta)S_{\bar A A}(-\theta)=1.
$$
Keeping the reversed species in this identity avoids an unstated parity assumption. For the particular amplitudes derived below, crossing and analytic <unitarity> also imply $S_{\bar A A}=S_{A\bar A}$. <Hermitian analyticity of a two-particle S-matrix> states $S_{ab}(\theta)^*=S_{ba}(-\theta^*)$ and, together with analytic <unitarity>, gives $|S_{ab}(\theta)|=1$ for real <rapidity> differences. <Crossing symmetry> analytically turns an incoming particle into an outgoing <antiparticle>, relating the two channels by $S_{A\bar A}(\theta)=S_{AA}(i\pi-\theta)$, with the reverse relation obtained by crossing again. The shift $i\pi$ follows from the sign reversal of the two-momentum $p(\theta)=(m\cosh\theta,m\sinh\theta)$.
Put $u=\lambda/2$. The given amplitude is the <unit-modulus hyperbolic scattering block> $S_u(\theta)=\sinh[(\theta+iu)/2]/\sinh[(\theta-iu)/2]$. Its numerator and denominator interchange under $\theta\mapsto-\theta$, proving analytic <unitarity>. For real $u$ it also obeys <Hermitian analyticity of a two-particle S-matrix>. Applying <crossing symmetry> gives
$$
\boxed{S_{A\bar A}(\theta)=\frac{\cosh[(\theta-iu)/2]}{\cosh[(\theta+iu)/2]}.}
$$
This crossed amplitude also has unit modulus on the real axis. Thus both requested channel constraints hold.
The bound-state conclusion needs a coupling range. In the fundamental attractive range $0<u<\pi$, equivalently $0<\lambda<2\pi$, the denominator has a simple <pole> at $\theta=iu$ inside the <physical rapidity strip>. Its <residue> is
$$
\operatorname*{Res}_{\theta=iu}S_{AA}(\theta)=2i\sin u,
$$
with positive imaginary coefficient. Under the usual one-particle interpretation of this direct-channel <bound-state pole>, it couples two charge-$+1$ particles to a new charge-$+2$ particle $B$. In the center-of-mass frame its constituents have analytically continued <rapidities> $\pm iu/2$, and their total two-momentum is $(2m\cos(u/2),0)$. Thus the <relativistic bound-state mass from a rapidity pole> gives
$$
\boxed{Q_B=+2,\qquad m_B=2m\cos\frac u2=2m\cos\frac\lambda4.}
$$
It is positive and less than the two-particle threshold $2m$. Without the coupling qualification the requested deduction is false: at $\lambda=0$ the amplitude is identically one after removing the apparent $0/0$ at the origin. A free massive complex <scalar field> has this diagonal amplitude, charge-$\pm1$ particles and no isolated charge-$+2$ <bound state>. At $u=\pi$ the amplitude similarly becomes constant $-1$ and the apparent boundary pole cancels. Neither endpoint supplies the claimed particle.
For <bound-state fusion of factorized S-matrices>, represent $B$ as the residue of $A(\theta_B+iu/2)A(\theta_B-iu/2)$ at its bound-state separation. Move a third $A(\theta_A)$ through both constituents using the <Faddeev-Zamolodchikov algebra>, then take the same residue. The bound-state normalization occurs on both sides and cancels. Consequently <bootstrap fusion> gives
$$
S_{BA}(\theta)=S_{AA}(\theta+iu/2)S_{AA}(\theta-iu/2)=\boxed{\frac{\sinh(\theta/2+3iu/4)\sinh(\theta/2+iu/4)}{\sinh(\theta/2-iu/4)\sinh(\theta/2-3iu/4)}},
$$
where $\theta=\theta_B-\theta_A$. This has analytic <unitarity> as a product of two shifted blocks. Its <poles> occur at $\theta=iu/2$ and $\theta=3iu/2$, modulo $2\pi i$; numerator zeroes occur at the corresponding negative positions, subject to cancellations at special couplings.
The nearer <pole>, $\theta=iu/2$, has <residue> $-2i\sin u$. It is a <crossed-channel pole in diagonal factorized scattering>, not a new direct-channel charge-$+3$ state. Indeed, the exchanged momentum has invariant
$$
t=m_B^2+m^2-2m_Bm\cos\frac u2=m^2,
$$
using $m_B=2m\cos(u/2)$. The exchanged particle is therefore the already present $A$, with the appropriate charge flow at the crossed vertex. Equivalently, <crossing symmetry> puts a positive-residue direct pole of $S_{B\bar A}$ at $i(\pi-u/2)$, corresponding to $B+\bar A\longrightarrow A$. This distinction avoids assigning an extra mass by applying the direct-channel formula to every <pole>.
The farther <pole>, $\theta=3iu/2$, lies in the <physical rapidity strip> only for $0<u<2\pi/3$, equivalently $0<\lambda<4\pi/3$. Its <residue> is
$$
2i\sin u\,\frac{\sin(3u/2)}{\sin(u/2)},
$$
which has positive imaginary coefficient in that range. It gives a charge-$+3$ <bound state> $C$. The <relativistic bound-state mass from a rapidity pole> now yields
$$
m_C^2=m_B^2+m^2+2m_Bm\cos\frac{3u}{2}=m^2(1+2\cos u)^2.
$$
The positive root in the admitted range is
$$
\boxed{Q_C=+3,\qquad m_C=m(1+2\cos u)=m\frac{\sin(3u/2)}{\sin(u/2)}=m(1+2\cos(\lambda/2)),\quad0<\lambda<\frac{4\pi}{3}.}
$$
This is the <three-particle bound state from equal-mass fusion>; in its own rest frame the constituent <rapidities> are $iu,0,-iu$. At $u=2\pi/3$ the farther apparent pole cancels because its numerator also vanishes; for $2\pi/3<u<\pi$ it lies outside the <physical rapidity strip> and does not require a new charge-$+3$ particle. Charge-conjugate partners carry charges $-2$ and $-3$ where the corresponding states exist. \b[Fusion distinguishes an existing crossed-channel particle from a genuinely new direct-channel <bound state>, and the latter requires the stated smaller coupling range.]
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