TITLE: Does *finitely many* include the option *none*?
QUESTION [10 upvotes]: Does finitely many include the option none?
Say I have a sequence $(x_n)$ and I want to say that there can only be $0$ or $n\in \mathbb N$ non-zero terms. Can I say that the sequence has finitely many non-zero terms?
Thanks.

REPLY [2 votes]: Gerd Faltings showed that $a^{n}+b^{n}=c^{n}$ has finitely many solutions $(a,b,c)$ for any $n>2$. 10 years later, Andrew Wiles showed that it had none. Faltings is not known to have contradicted.