shall also denote, according to ref. [@next], as $s(2)$: $$Q^{\pm2}_{\epsilon}=0,\quad \{Q^{+}_{\epsilon},Q^{-}_{\epsilon}\}
=H_{\epsilon},\quad
[Q^{\pm}_{\epsilon},H_{\epsilon}]=0.
\label{n2s2}$$ Note that the hermitian supercharge operators $Q^{1,2}_{\epsilon}$, $$Q^{\pm}_{\epsilon}=\frac{1}{2}(Q^{1}_{\epsilon}\pm iQ^{2}_{\epsilon}),\quad
\{Q^{i}_{\epsilon},Q^{j}_{\epsilon}\}=2\delta^{ij}H_{\epsilon},$$ have the following compact form: $$Q^{1}_{\epsilon}=\frac{1}{\sqrt{2}}(q+i\epsilon pK),\quad
Q^{2}_{\epsilon}=\frac{1}{\sqrt{2}}(p-i\epsilon qK)=
-i\epsilon Q^{1}_{\epsilon}K
\label{q12}$$ in terms of the coordinate $q$ and momentum $p$ operators of the deformed bosonic oscillator introduced by eq. (\[apq\]).

Consider the spectrum of the supersymmetric Hamiltonian (\[hexp1\]). Using the expression for the number operator given by eq. (\[ndef\]), we find that in the case when $\epsilon=-$, the states $|n>$ are the eigenstates of the operator $H_{-}$ with the eigenvalues $$E_{n}^{-}=2[n/2]+1+\nu,$$ where $[n/2]$ means the integer part of $n/2$. Therefore, for $\epsilon=-$ we have the case of spontaneously broken supersymmetry, with $E_{n}^{-}>0$ for all $n$ due to restriction (\[restr\]). All the states $|n>$ and $|n+1>$, $n=2k$, $k=0,1,\ldots$, are paired here in supermultiplets. For $\epsilon=+$, we have the case of exact supersymmetry characterized by the spectrum $$E_{n}^{+}=2[(n+1)/2],$$ i.e. here the vacuum state with $E_{0}^{+}=0$ is a supersymmetry singlet, whereas $E_{n}^{+}=E_{n+1}^{+}>0$ for $n=2k+1,
k=0,1,\ldots$. Hence, we have demonstrated that one can realize both cases of the spontaneously broken and exact $N=2$ supersymmetries with the help of only one bosonic oscillator, and in the former case the scale 