For $v=44$, points either lie on four 8-lines (types A), or one 8-line
(types B).  If there are $M$ type A points, then divisibility
conditions for 8-lines imply that $M\equiv4$ mod 8, say $M=4+8m$, and
$N_8=7+3m$; if there are at most 40 type B points, and at least 7
8-lines, then some 8-line contains at least 3 type A points, and so
$N_8\geq 3*(4-1)+1=10$. Repeating the above arguement with 40, 7, 3,
$N_8\geq10$, replaced by 32, 10, 5 $N_8\geq16$; by 16, 16, 7,
$N_8\geq22$; by 0, 22, 8, $N_8\geq25$ eliminates all possible point
distributions.

For $v=45$, points either lie on five 8-lines (types A), or two 8-line
(types B).  If there are $M$ type A points, then divisibility
conditions for 8-lines imply that $M\equiv2$ mod 8, say $M=2+8m$, and
$N_8=12+3m$; if there are at most 35 type B points, and at least 15
8-lines, then some 8-line contains at least 4 type A points, and so
$N_8\geq 4*(5-1)+4*(2-1)+1=21$. Repeating the above arguement with 19,
21, 7, $N_8\geq30$ eliminates all possible point distributions except
2 type A points, with $N_8=12$.  However, in this case we cannot have
two type A points on a 8-line as this implies $N_8\geq 15$; also, the
only possible A types are $4^1 7^1 8^5$ and $4^3 8^5$, but the first
of these only has degree 7, and cannot lie on all 8-lines, so is not
viable.  Thus, both type A points must meet on 4-lines, and that
4-line intersects $2*5+2*2$ 8-lines, so $N_8=12$ eliminates this
possibility.

For $v=47$, points either lie on four 8-lines (types A), or one 8-line
(types B).  If there are $M$ type A points, then divisibility
conditions for 8-lines imply that $M\equiv3$ mod 8, say $M=3+8m$, and
$N_8=7+3m$; if there are at most 36 type B points, and at least 10
8-lines, then some 8-line contains at least 5 type A points, and so
$N_8\geq 5*(4-1)+1=16$. Repeating the above arguement with 20, 16, 7,
$N_8\geq22$, and with 4, 22, 8 $N_8\geq25$ eliminates all possible
point distributions except $M=3$ and $N_8=7$, but then these three
type A points have 12 incidences on the 7 8-lines, which causes too
many pairs, so there is no viable distribution.


-- Malcolm Greig
Jan. 1998.