Many of them were already known to Gauss.
The proof makes use of many of the ideas of the general case, but in a simpler setting.
Thus $G$ has 10 normal subgroups and as many non-normal ones.
Consequently, $H$ is a free $R$-module on as many generators as there are path components of $X$.
Therefore, $A$ has two elements too many. [Or: $A$ has two too many elements.]
Then $A$ has three times as many elements as $B$ has.
It meets only countably many of the $Y_i$.
a sequence with only finitely many terms nonzero
To compute how many such solutions there are, observe ......
How many of them are convex?
How many such expressions are there?
How many entries are there in this section?
How many multiplications are done on average?
How many zeros can $f$ have in the disc $D$?