I have an homework to do and i have no idea where to start. The question is:
"Show that a narrow-sense binary BCH-code of length $ n=2^m-1 $ and designed distance $ 2t+1 $ has minimum distance $ 2t+1 $, provided that $$ \sum_{i=0}^{t+1}\binom{2^{m}-1}{i}>2^{mt} $$"
Please help me solve this question.
Completing the hint by Dilip Sarwate to an answer.
Let $S$ be the set of all binary vectors of length $n$ and weight at most $t+1$. Then $$ |S|=\sum_{i=0}^{t+1}{2^m-1\choose i}. $$ Let $H$ be the parity check matrix. If $x$ is any binary vector of length $n$, then the syndrome $Hx^T$ has $mt$ binary components. The inequality $|S|>2^{mt}$ tells us that there are (at least) two distinct vectors $x_1$ and $x_2$ in $S$ such that their syndromes are equal, i.e. $Hx_1^T=Hx_2^T$. Then $H(x_1-x_2)^T=0$, so $x_1-x_2$ is a codeword.
But $x_1-x_2$ has weight at most $2(t+1)$, so the minimum distance of this code cannot be higher than $2t+2$. But we also know (have you covered this?) that the minimum distance of a narrow sense binary BCH-code cannot be even. This follows for example from the fact extended binary BCH-codes have a transitive automorphism group (translations by an element of $GF(2^m)$). Therefore the extended BCH-code has words of minimum weight with one of the 1s at position zero. This is the position that is deleted when going back from the extended code to the unextended version.