 Solution

Let $k$ count parallel bonds. Then $N-k$ bonds are antiparallel and the energy isThe periodic boundary conditions imposeHence $N-k$ must be even. For odd $N$, $k$ is odd, so its minimum is one: one bond must be parallel. All other bonds alternate. Choosing that bond in $N$ ways and then choosing either value of $S_1$ determines every Ising spin, givingThis is an odd antiferromagnetic Ising ring: geometrical frustration forces one defect, whose location is unrestricted. The ground-state degeneracy includes both its possible locations and global spin inversion symmetry. The formula also includes the self-bond case $N=1$ under the stated Hamiltonian convention.