Single-error test from two binary BCH syndromes ($S_1={\alpha }^j,S_3={\alpha }^{3j}=S_1^3$)

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 syndromesThus $S_1\ne 0$ 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.