A non-coprime phase gives a proper order divisor

Direct multiplication gives $2^1\equiv 2$, $2^2\equiv 4$, $2^3\equiv 3$, and $2^4\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}5)$, with no earlier power equal to $1$; thus the order is exactly $4$. The specified register size gives $n=\lceil {\log }_{2}5\rceil =3$, $m=2n+3=9$, and $Q=2^9=512$.

For $k=2$, the phase reduces to $1\mathrm{/}2$, so the exact grid label is $y=512\mathrm{/}2=256$. Applying [F3] with reduced numerator $1$, denominator $2$, and $N=5$ gives reconstructed denominator $q=2$. The verification test is $2^2=4\equiv ̸1(\mathrm{m}\mathrm{o}\mathrm{d}5)$, hence [F1] rejects this candidate.

The four vectors $\mathrm{\mid }{\psi }_k⟩=\frac{1}{2}{\sum }_{s=0}^3{i}^{-ks}\mathrm{\mid }{2}^s<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>5⟩$ are eigenvectors with phases $k\mathrm{/}4$: multiplication by $2$ cycles $1,2,4,3$, and shifting the sum multiplies it by $i^k$. The four-term geometric sum gives $\mathrm{\mid }1⟩=\frac{1}{2}{\sum }_{k=0}^3\mathrm{\mid }{\psi }_k⟩$, so [F2] assigns weight $1\mathrm{/}4$ to each phase and returns its exact grid label. For $k=1$, $y=512\mathrm{/}4=128$; [F3] reconstructs denominator $4$, and $2^4=16\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}5)$, so [F1] retains it. Thus both the rejected $k=2$ outcome and a successful $k=1$ outcome have positive probability.

By [F4], every retained denominator is a multiple of $4$. The bad denominator $2$ is rejected by step 2.1, while the denominator-$4$ sample in step 2.2 is retained; [F5] then makes the minimum retained value $4$. This example verifies that a reconstructed divisor is never silently reported as the exact order.