Write $a=|\alpha {|}^2$, $b=|\beta {|}^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party giveswhere $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate eigenspace, rather than measuring her four bits separately. Her projective measurement has projectors ${\Pi }_k={\sum }_{z(x)=k}|x⟩⟨x|$, and the Born rule givesAlice sends $k$ to Bob. Up to the common phase of ${\alpha }^k{\beta }^{4-k}$, their conditional pure state isThis is Schmidt projection entanglement concentration: its equal Schmidt coefficients give a maximally entangled state of Schmidt rank $d_k$.

For $k=0$ or $4$, $d_k=1$, and the branch is a product state. No Bell pair can be extracted using local operations and classical communication. The respective probabilities are $b^4$ and $a^4$.

For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a local unitary operation extending the basis relabellingSuch a unitary exists because these are four distinct orthonormal basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leavesThus these branches give exactly two Bell pairs, with probabilities $4a{b}^3$ and $4a^{3}b$, and no further measurement in quantum mechanics is needed.

For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local POVM gives two Bell pairs with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the Kraus operatorsOn the six-dimensional support, each label is retained by two operators, so ${\sum }_{r=0}^2{K}_r^{†}{K}_r={\Pi }_2$. Add $K_{\perp }=I-{\Pi }_2$ to define a complete measurement in quantum mechanics on Alice's four qubits; that extra outcome has probability zero on this branch. Each nonzero outcome has conditional probability $(1/2)(4/6)=1/3$, and gives the normalized stateAlice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00⟩,|01⟩,|10⟩,|11⟩$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two Bell pairs. The joint probability of $(k,r)=(2,r)$ is $p_2/3=2a^2{b}^2$ for each $r$.

For an explicit local implementation of this POVM, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a deterministic reduction of maximally entangled Schmidt rank from six to four.

The complete output distribution for this strategy is consequentlyThe mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has Schmidt rank at most six, whereas three independent Bell pairs require Schmidt rank eight. In any refined LOCC branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its rank.

The EPR criterion of reality says that if a physical quantity can be predicted with certainty without disturbing the system, then it represents an element of physical reality. For separated particles, the EPR locality assumption regards choices of measurements on the other particles as nondisturbing. The perfect correlations below therefore imply predetermined local values for both $X={\sigma }_x$ and $Y={\sigma }_y$, independent of which distant measurements are chosen.

Put $|0⟩=|\uparrow ⟩$ and $|1⟩=|\downarrow ⟩$. The Pauli matrices satisfyFor three particles, each operator with one $X$ and two $Y$ takes $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$, whereas $X\otimes X\otimes X$ interchanges the two strings without a minus sign. Thus the GHZ state in the question has the certain outcomesEach local $X$ or $Y$ value can be predicted from measurements at the other two sites in one of these contexts. If the EPR criterion of reality and locality assign values $x_j,y_j\in \{\pm 1\}$, the first three equalities requireMultiplying these ordinary numbers and using $y_j^2=1$ yields $x_1{x}_2{x}_3=1$, contradicting the certain quantum outcome $-1$. This derives the GHZ theorem explicitly; the multiplication is of hypothetical local values, not of mutually incompatible local observables.

For general $N$, let $O_j=X_j{⨂}_{k\ne j}{Y}_k$, with the tensor factors ordered by their site labels, and let $X_N=X^{\otimes N}$. Directly,For the difference of the two strings to be an eigenstate, these phases must agree, which happens exactly when $N-1$ is even. For odd $N$ the common phase is $(-1{)}^{(N-1)/2}$, and interchanging the strings negates their difference. ConsequentlyFor even $N$, $X_N$ still has eigenvalue $-1$, but every $O_j$ takes the difference state to a phase times the sum state, which is orthogonal to it. Thus it is not an eigenstate of any $O_j$; those operators each have outcomes $\pm 1$ with equal probabilities. Among the stated $N\ge 4$, simultaneous eigenstates therefore occur precisely at $N=5,7,9,\dots$.

For odd $N$, certainty of each $O_j$ again permits remote prediction of both local $X$ and $Y$. Writing ${\lambda }_N=(-1{)}^{(N+1)/2}$, the local hidden-variable theory would require $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ for all $j$. Their product ishence ${\prod }_j{x}_j={\lambda }_N$, since $N-1$ is even and $N$ is odd. Quantum theory requires the same product to be $-1$. The contradiction occurs when ${\lambda }_N=+1$, giving the infinite familyFor $N\equiv 1(\mathrm{mod}4)$, these particular certain product constraints are consistent; for example $x_j=-1$, $y_j=+1$ satisfies them. This distinguishes the GHZ contradiction for N congruent to 3 modulo 4 from the larger set of odd $N$ for which simultaneous eigenstates exist.