How many integer solutions of "a" exist for this equation? $a\times b = 8 \times (a + b)$
I have used Wolfram Alpha and it has given me 14 integer solutions.
But how can we find those solutions ? Which method ?
edit: integer solutions are as following:
$a = [-56, -24, -8, 0, 4, 6, 7, 9, 10, 12, 16, 24, 40, 72]$
$ b = [7, 6, 4, 0, -8, -24, -56, 72, 40, 24, 16, 12, 10, 9]$

So, $$a=\frac{8b}{b-8}=\frac{8(b-8)+64}{b-8}=8+\frac{64}{b-8}\iff (b-8)\mid64$$
Now, $64=2^6$ has $6+1=7$ positive divisors 
So, we have $7$ solutions if we consider the only the positive values of $b-8$.
So, if we include all the non-zero integers, we shall have $2\cdot7=14$ divisors.
Clearly, $b-8\ne0$ else $a$ will be infinite.
So, the values of $b-8$ are $\pm 2^i$ for integer $i\ge0$ and $\le6$
Hence, 
if $i=0,b-8=\pm1,b=9,7;$
if $i=1,b-8=\pm2,b=10,6;$
if $i=2,b-8=\pm4,b=12,4;$
if $i=3,b-8=\pm8,b=16,0;$
if $i=4,b-8=\pm16,b=24,-20;$
if $i=5,b-8=\pm32,b=40,-24;$
if $i=6,b-8=\pm64,b=72,-56;$
