# 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:

$a = [-56, -24, -8, 0, 4, 6, 7, 9, 10, 12, 16, 24, 40, 72]$

$b = [7, 6, 4, 0, -8, -24, -56, 72, 40, 24, 16, 12, 10, 9]$

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Hint: $$ab-8(a+b)=0 \iff (a-8)(b-8)=64$$

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the amount of integers can be found this way but finding the exact values of those integers is still bugging me. –  Selim Arikan Feb 18 '13 at 13:56
@SelimArikan Use factorization of 64. For example, $$(a-8)(b-8)=32\cdot 2$$ So one proposed solution of this equation is $a-8=32$, $b-8=2$, so $a=40, b=10$. –  tetori Feb 19 '13 at 1:51
thank you for your explanation –  Selim Arikan Feb 19 '13 at 7:25

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 $i=6,b-8=\pm64,b=72,-56;$

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But there are only three negative solutions for a, [-56, -24, -8] and how to find those values ? –  Selim Arikan Feb 18 '13 at 12:02
@SelimArikan, observe that we have $7$ negative and $7$ positive values for $b-8$. Sorry for the delay. –  lab bhattacharjee Feb 18 '13 at 15:00
thank you for your explanation –  Selim Arikan Feb 19 '13 at 7:24