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7
votes
If $x_{n+1}= \frac{nx_{n}^2+1}{n+1}$ then $x_{n}=1$
I suspect the answer is no. First rewrite $x_n=y_n+1$, then the recursion becomes
$(n+1)y_{n+1}=ny_n(y_n+2)=(y_n+2)(y_{n-1}+2)\cdots (y_2+2) (y_1+2)y_1$
and for the integrality of $y_{n+1}$ it is su …
3
votes
Ways to prove the fundamental theorem of algebra
At the risk of being highly downvoted, I can't resist reposting my comment
to Andrew L's answer (or rather, question) below:
is there a purely algebraic proof that for any non constant $P$ in $\mathb …