= Single-error test from two binary BCH syndromes
{title2=$S_1=\alpha^j,\quad S_3=\alpha^{3j}=S_1^3$}
Let a binary <cyclic code> of length $n$ have <defining zeros of a cyclic code> $\alpha$ and $\alpha^3$, with $\alpha$ of <multiplicative order> $n$. A single error in position $j$ has error <polynomial> $X^j$ and <syndromes>
$$
S_1=\alpha^j,\qquad S_3=\alpha^{3j}=S_1^3.
$$
Thus $S_1\ne0$ and $S_3=S_1^3$ are necessary for a single-error pattern. Search the $n$ distinct powers of $\alpha$ for $S_1$; if it is absent, this is not a single-error <syndrome>. Otherwise flip the unique coordinate $j$ and test membership in the full <cyclic code>, for example by divisibility by its entire <generator polynomial of a cyclic code>.
Checking only the two roots is sufficient only when they generate all defining checks, including their field conjugates. A <cyclic code> with additional defining roots requires those checks too. If the full membership test succeeds and the <minimum Hamming distance> is at least five, the corrected <codeword> is the unique <codeword> within two errors of the received word: two such <codewords> would be at <Hamming distance> at most four. The two-<syndrome> relation alone does not establish membership.
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