= Solution
Use the minimal four-dimensional <Super-Poincaré algebra>, with $\{Q_\alpha,Q_\beta\}=0$, $\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\nu_{\alpha\dot\beta}P_\nu$ and commuting translations. <Lorentz covariance> and closure on the existing <supercharges> permit only
$$
[P^\mu,Q_\alpha]=A(\sigma^\mu)_{\alpha\dot\gamma}\bar Q^{\dot\gamma}.
$$
The <graded Jacobi identity> for $P^\mu,Q_\alpha,Q_\beta$ therefore gives
$$
0=\{[P^\mu,Q_\alpha],Q_\beta\}+\{Q_\alpha,[P^\mu,Q_\beta]\}
=2A\left(\sigma^\mu_{\alpha\dot\gamma}\epsilon^{\dot\gamma\dot\delta}\sigma^\nu_{\beta\dot\delta}+\sigma^\mu_{\beta\dot\gamma}\epsilon^{\dot\gamma\dot\delta}\sigma^\nu_{\alpha\dot\delta}\right)P_\nu.
$$
For example, with $\sigma^0=I$, $\mu=0$ and $\alpha=\beta=1$, the coefficient is a nonzero multiple of $A(P_1-iP_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus \b[every supercharge commutes with four-momentum]:
$$
\boxed{[Q_\alpha,P^\mu]=0.}
$$
Consequently $[Q_\alpha,P^2]=0$. If $|u\rangle$ has a particular <mass-shell condition>, the nonzero state $Q_\alpha|u\rangle$ has the same one. This gives <supersymmetric mass degeneracy> within an unbroken physical <supermultiplet>. The qualification matters: after spontaneous <supersymmetry breaking>, the chosen vacuum is not annihilated by the <supercharges>, and its one-particle excitations need not constitute degenerate <supermultiplets>.
For the <O'Raifeartaigh model>, write $X=\phi_1$, $Y=\phi_2$, $Z=\phi_3$. Field phases allow $g,M,m$ to be positive. The canonical <Kähler potential> and elimination of the <auxiliary fields> give the <F-term scalar potential>
$$
V=g^2|Z^2-m^2|^2+M^2|Z|^2+|2gXZ+MY|^2.
$$
A <supersymmetric vacuum> would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-gm^2$ could not vanish. Thus <F-flatness> is impossible. More precisely, with $Z=u+iv$,
$$
V=g^2m^4+(M^2-2g^2m^2)u^2+(M^2+2g^2m^2)v^2+g^2(u^2+v^2)^2+|2gXZ+MY|^2.
$$
The stated strict inequality makes every nonconstant term nonnegative. \b[The entire classical vacuum family is]
$$
\boxed{Z=Y=0,\qquad X=x\text{ arbitrary},\qquad V_0=g^2m^4>0.}
$$
Here $F_X=-\overline{W_X}=gm^2$, while $F_Y=F_Z=0$. The <Weyl spinor> $\psi_X$ is the massless <goldstino>, and the <complex scalar field> $X$ is a <pseudomodulus> with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.
At the representative vacuum $x=0$, the <chiral-superfield fermion mass matrix> is
$$
m_F=\begin{pmatrix}0&0&0\\0&0&M\\0&M&0\end{pmatrix}.
$$
Its physical masses are its <singular values>: $0,M,M$. The two massive <Weyl spinors> can be combined into one massive <Dirac spinor>. For $Y=(y_R+iy_I)/\sqrt2$, $Z=(z_R+iz_I)/\sqrt2$, the real scalar squared masses are
$$
\boxed{m_{X_R}^2=m_{X_I}^2=0,\quad m_{y_R}^2=m_{y_I}^2=M^2,\quad m_{z_R}^2=M^2-2g^2m^2,\quad m_{z_I}^2=M^2+2g^2m^2.}
$$
In particular, there is no <tachyon> under the stated inequality. The splitting of the $Z$ masses displays <supersymmetry breaking> directly.
For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the <pseudomodulus>. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses are
$$
f_\pm=M^2+2g^2|x|^2\pm2g|x|\sqrt{M^2+g^2|x|^2}.
$$
The real and imaginary scalar sectors have two-by-two <scalar squared-mass matrices>
$$
B_\epsilon=\begin{pmatrix}M^2&2gM|x|\\2gM|x|&M^2+4g^2|x|^2+2\epsilon g^2m^2\end{pmatrix},\qquad\epsilon=-1,+1,
$$
whose <eigenvalues> are
$$
b_{\epsilon,\pm}=M^2+2g^2|x|^2+\epsilon g^2m^2\pm\sqrt{(2g^2|x|^2+\epsilon g^2m^2)^2+4g^2M^2|x|^2}.
$$
Both <determinants> are $M^2(M^2+2\epsilon g^2m^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a <pseudomodulus>; these formulas describe the classical, tree-level spectrum requested here.
Define the physical mass <supertrace> by
$$
\operatorname{STr}\mathcal M^2=\sum_s(-1)^{2s}(2s+1)\operatorname{tr}\mathcal M_s^2.
$$
Each real scalar contributes once and each <Weyl spinor> twice with a minus sign. At $x=0$ this gives $4M^2-2(2M^2)=0$. At general $x$, $\sum_{\epsilon,\pm}b_{\epsilon,\pm}=4M^2+8g^2|x|^2=2(f_++f_-)$. Hence \b[the tree-level mass supertrace vanishes despite broken supersymmetry]:
$$
\boxed{\operatorname{STr}\mathcal M^2=0.}
$$
This is the <tree-level supertrace mass sum rule>, not a claim that the individual masses are equal.
For direct breaking in the <Minimal supersymmetric Standard Model>, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant <superpotential> term needs a gauge-singlet <chiral superfield>, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral <F-term> breaking, without <D-term> mass shifts, the <MSSM tree-level sfermion mass constraint> applies separately to conserved charge sectors. For one electron pair it gives $m_{\tilde e_1}^2+m_{\tilde e_2}^2=2m_e^2$: both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total <supertrace> alone. A separate <hidden supersymmetry-breaking sector> communicating effective <soft supersymmetry breaking> through loops or suppressed operators avoids the direct canonical tree-level obstruction.
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